The  MARTIN  ROWAN  CHAFFIN 
Collection  of  Pablic  School 


By  his  grandchildren  in  honor  of  M. 
R.  Chaffin,  who  taught  public  school 
in  Davie  and  Yadkin  counties  for  a 
number  of  years  beginning  in  1850,  and 
in  honor  of  his  father,  William  Owen 
Chaffin,  who  first  taught  a North 
Carolina  public  school  in  1843,  in 
Yadkin  county. 

For  the  especial  use  of  the  Department  of 
Education  and  of  the  Durham  county  and 
city  teachers. 


Text-Books 


PRESENTED  TO 


DA  TE 


,3> 


Digitized  by  the  Internet  Archive 
in  2016  with  funding  from 
Duke  University  Libraries 


https://archive.org/details/elementarycalcul01smit 


ELEMENTARY  CALCULUS 


A TEXT-BOOK  FOR  THE  USE  OF 
STUDENTS  IN  GENERAL  SCIENCE 


BY 

PERCEY  F.  SMITH,  Ph.D. 

PROFESSOR  OF  MATHEMATICS  IN  THE  SHEFFIELD  SCIENTIFIC  SCHOOL 
OF  YALE  UNIVERSITY 


NEW  YORK- : • CINCINNATI  • : • CHICAGO 

AMERICAN  BOOK  COMPANY 


Copyright,  1902  and  1903,  by 

PERCEY  F.  SMITH. 


EL.  CALC  SMITH. 


W.  P.  4 


£ * £ * 

si 7 

S bTL  £ 

\ 


PREFACE 


This  volume  has  been  written  in  response  to  the  un- 
mistakable and  growing  demand  for  a text-book  on  the 
Calculus  which  shall  present  in  a course  of  from  thirty- 
five  to  forty  exercises  the  fundamental  notions  of  this 
branch  of  mathematics.  In  American  technical  schools 
students  pursuing  courses  distinct  from  engineering 
branches  usually  terminate  their  mathematical  studies 
with  Plane  Analytic  Geometry.  But  in  view  of  the  recent 
remarkable  development  of  certain  of  the  general  sciences 
along  mathematical  lines,  such  a course  can  no  longer  be 
regarded  as  adequate.  Moreover,  there  can  be  no  differ- 
ence of  opinion  as  to  the  relative  advantage  to  the  student 
of  a knowledge  of  more  than  the  mere  elements  of  Ana- 
lytic Geometry  and  an  introductory  acquaintance  with  the 
Calculus.  It  is,  I think,  the  experience  of  every  teacher 
that  the  average  student  first  realizes  the  power  and  use 
of  mathematics  when  taught  to  solve  problems  in  maxima 
and  minima  by  means  of  the  methods  of  the  Differential 
Calculus.  Certainly  no  stronger  argument  can  be  adduced 
in  favor  of  an  adjustment  of  the  curriculum  which  shall 
include  this  branch  of  mathematics.  Such  a change  has 
been  effected  in  the  Sheffield  Scientific  School,  and  results 
abundantly  justify  the  step. 

For  the  general  student  in  our  colleges  who  elects  a 
year’s  work  in  mathematics  beyond  the  usually  required 


3 


4 


PREFACE 


Trigonometry,  the  most  satisfactory  course  would  seem  to 
be  one  in  which  the  time  is  equally  divided  between  Plane 
Analytic  Geometry  and  Calculus. 

In  writing  this  book  I have  everywhere  emphasized  the 
possibility  of  applications.  The  examples  have  been  care- 
fully selected  with  this  end  in  view.  The  first  chapter 
may  seem  long,  but  the  notion  of  limit  certainly  demands 
adequate  treatment.  While  an  elementary  text-book  offers 
no  excuse  for  employment  of  the  refinements  of  modern 
rigor,  I have  endeavored  to  avoid  positive  inaccuracies 
and  have  carefully  distinguished  between  demonstration 
and  illustration. 

I am  indebted  to  my  colleague,  Dr.  W.  A.  Granville,  for 
many  helpful  suggestions. 

PERCEY  F.  SMITH. 

Sheffield  Scientific  School. 


CONTENTS 

CHAPTER  PAGE 

I.  Functions  and  Limits  . 7 

II.  Differentiation  ........  23 

III.  Applications 51 

IV.  Integration 70 

V.  Partial  Derivatives . 84 

VI.  Additional  Examples  .......  90 


5 


349462 


ELEMENTARY  CALCULUS 

CHAPTER  I 
FUNCTIONS  AND  LIMITS 

1.  Continuous  Variation.  In  this  book  we  are  concerned 
with  real  numbers  only.  Geometrically,  such  numbers  may 
be  conveniently  represented  by  points  of  a scale  (Fig.  i). 

etc. -5  — i —8  Eg  r£  o 1 2 3 t 5 6 7 etc. 

Fig.  i 

Then  to  every  real  number  corresponds  one  point  of  the 
scale,  and  only  one ; conversely,  every  point  of  the  scale 
represents  a real  number.  Any  segment  of  the  scale, 
however  small,  represents  indefinitely  many  numbers.  We 
speak  indifferently  of  the  number  a and  the  point  a of  the 
scale. 

A variable  x is  said  to  vary  contmuously  between  the 
numbers  a and  b when  it  assumes  values  corresponding  to 
every  point  of  the  segment  ab. 

2.  Functions.  The  problems  arising  in  Elementary 
Calculus  involve  in  general  two  variables  in  such  a way 
that  the  value  of  one  variable  can  be  calculated  as  soon  as 
a value  is  assumed  for  the  other.  Thus,  in  Geometry,  the 
student  has  an  illustration  in  the  area  and  radius  of  a 
circle,  two  variables  such  that  the  area  A can  be  calculated 
when  we  know  the  radius  r from  the  formula  A — n tA. 


7 


8 


FUNCTIONS  AND  LIMITS 


Definition.  A variable  is  said  to  be  a function  of  a second 
variable  when  its  value  depends  upon  the  value  of  the  sec- 
ond variable  and  can  be  calculated  when  the  value  of  the 
second  variable  is  assumed. 

The  first  variable  is  called  the  dependent  variable , and 
the  second  the  independent  variable. 

For  example,  the  equations 

y = -r2,  y = sinx,  y = log10(.r2  - i) 

state  that  y is  a function  of  x.  In  the  first  two  cases,  y 
may  be  calculated  for  any  value  of  x ; in  the  last  case,  how- 
ever, x is  restricted  to  values  numerically  greater  than  I, 
since  the  logarithms  of  negative  numbers  cannot  be  found. 
In  the  first  two  cases,  then,  we  say  that  the  dependent 
variable  (or  the  function)  is  defined  for  every  value  of  x, 
and  in  the  last  the  function  is  defined  only  when  x exceeds 
i numerically. 

A function  is  defined  for  a value  of  the  variable  when  its 
value  can  be  calculated  for  that  value  of  the  variable. 

Elementary  Functions.  Power  Function:  xm,  m any 
positive  integer. 

Logarithmic  Function:  loga;r,  a>  o;  this  function  is 
defined  only  for  x>o. 

Exponential  Function : ax,  a > o,  He.  the  exponent  is  a 
variable,  the  number  a being  a constant. 

Circular  Functions  :*  sin  x,  cos  jr,  tan  .r,  etc.,  i.e.  involving 
the  six  trigonometric  functions. 

* So  called  from  the  use  of  the  circle  in  their  definition, 

e.g.  in  Fig.  2,  sin  A OP  = FL  = .\fp  jf  ft  js  the  unit  of 
R 

linear  measure.  Hereafter,  angles  will  always  be  meas- 
ured in  circular  measure , i.e.  x = arc—  = 'LL.  In  the 

radius  R 

FlG.  2 unit  circle,  R = I,  x = arc  AP. 


FUNCTIONS  AND  LIMITS 


9 


Inverse  Circular  Functions:  arc  sin  x,  arc  tan  x,  etc.,  i.e. 
the  “arc  whose  sine  is  x,”  “arc  whose  tangent  is  x,”  etc. 
In  the  unit  circle  (see  Fig.  2)  R = 1 ; if  x=MP,  then 
arc  sin  x = arc  AP. 

One  thing  is  peculiar  here.  Assuming  any  value  of  x 
not  exceeding  1 numerically , arc  sin  x may  be  calculated, 
but  the  number  of  answers  is  always  indefinitely  great. 
For  not  only  is 

arc  sin  MP  = arc  AP, 

but  also  equal  to  any  number  of  circumferences  + arc  AP, 
i.e.  arc  sin  MP  = arc  AP  + 2 ml, 

where  n is  any  integer. 

For  this  reason  the  inverse  circular  functions  are  called 
many-valued  functions.  For  definiteness  we  may  always 
take  the  least  positive  arc. 

3.  Functional  Notation.  As  general  symbols  for  func- 
tions of  variables  we  use  the  notation 

f(x),  0(y),  (f>(r),  etc., 

(read  f function  of  x,  theta  function  of  y,  phi  function  of 
r,  etc.). 

We  mean  by  this  that  f(x)  is  a variable  whose  value  de- 
pends upon  x,  and  can  be  found  when  a value  is  assumed 
for  .r.  The  notation  is  extremely  convenient,  for  it  enables 
us  to  indicate  the  value  of  the  function  corresponding  to 
any  value  of  the  variable  for  which  the  function  is 
defined. 

Thus  f(a)  represents  the  value  of  f(x)  for  x = a,  6(0) 
the  value  of  6 ( y)  for  y = o,  <£(^)  the  value  of  <f>(r ) for 
r = 1,  etc. 


IO 


FUNCTIONS  AND  LIMITS 


EXERCISE  1 

1.  For  what  values  of  the  variable  are  the  following  functions 
defined  ? 

(a)  A?is.  For  every  value  except  ;r  = o,  since  - cannot  be 

x calculated.*  0 

(b)  vhr2  — 6x.  Since  x-  — 6 x or  x(x  — 6)  must  not  be  negative, 
x and  x — 6 must  always  have  the  same  signs. 

Ans.  For  every  value  except  those  between  o and  6. 

0)  y/y  - y2 ; (d)  vTo;  ( e ) arc  sin  x; 

(/)  arc  sec  x;  (g)  sin  Vi  + x;  ( h ) log  tan  x. 

2.  Given  f(x)  = x3  — 7 x2+  16  x—  12,  show  that  f(j2.)  =0,  /($)  —o. 
Does  f(x)  vanish  for  any  other  value  of  x ? 

3.  Given  f(x)  = logx;  show  that 

/O')  +/OO  =/(*/)• 

4.  Given  $ ( x ) = az  ; show  that 

O')  <£(/)  = <£  O' + /)• 

5.  Given  0 (x)  = cosjt; 

then  6 (x)  + 6 ( y ) = cos  x + cos y. 

From  Trigonometry,  we  know  that 

cos  .r  + cos/  = 2 cos  £ (x  + /)  cos  £ (jr  — /)  ; 

4.  Graph  of  a Function.  After  determining  for  what 
values  of  the  variable  a given  function  is  defined,  it  is  im- 
portant to  know  in  what  manner  the  value  of  the  function 


* The  student  should  observe  that  the  four  fundamental  operations  of  arithmetic, 
addition,  subtraction,  multiplication,  and  division,  when  performed  with  real  num- 
bers, give  real  numbers,  with  the  single  exception  that  division  by  zero  is  cxchided. 


FUNCTIONS  AND  LIMITS 


II 


changes  with  the  variable.  Geometrically  this  is  accom- 
plished by  drawing  the  graph  of  the  function , which  is 
thus  defined : 

The  graph  of  a function  is  the  curve  passing  through  all 
points  whose  abscissas  are  the  values  of  the  variable  and 
ordinates  the  corresponding  values  of  the  function. 

In  the  language  of  Analytic  Geometry  the  graph 
of  a function  fix)  is  the  locus  of  the  equation 

y =/(*)• 


— l-  - 


Fig.  3 

By  carefully  drawing  the  graph  of  a func- 
tion a good  idea  is  obtained  of  the  behavior 
of  the  function  as  the  variable  changes.  For  ex- 
ample, the  graph  of  log3  -r,  i.e.  the  locus  of  the 
equation  * 

y = log3x, 

is  drawn  in  Fig.  3. 

Here  we  see  the  following  facts  clearly  pictured 
to  the  eye. 

(a)  For  x=  i,  log3^=  log3  1=0. 

( b ) For  x>  1,  logger  is  positive  and  increases  as  jv  in- 

creases. 

* The  values  ofy  are  found  from  the  formula  proven  in  the  theory  of  logarithms, 
logio  x 


logs  * = 


logio  3 


12 


FUNCTIONS  AND  LIMITS 


( c ) For  x<  i,  logg.tr  is  negative  and  increases  indefi- 

nitely in  numerical  value  as  x diminishes. 

( d ) For  .r  = o,  log3  .tr  is  not  defined,  since  the  logarithm 

of  zero  cannot  be  calculated. 


The  graph  of  the  general  logarithmic  function  loga  ;tr 
may  be  drawn  by  merely  changing  the  ordinates  in  Fig.  3 
in  the  constant  ratio  1 -5-  log3  a. 


Graphs:  (a)  Of  x2. 


(b)  Of  x*. 


Y 


The  graph  of  xm  has  the  appearance  of  (a)  or  (b)  according  as  m 
is  even  or  odd. 


00  Of  log3  a-. 


(d)  Of  31. 


1 

/ 

T 

I 

0 

/ 

r 

(T 

\-i- 

1 

1 

3 

jLd-)- 

1 1 

“/ 

F 

Trl 

Since  if  we  set  y = log3a-,  then  x=y,  the  graph  in  ( c ) has  the 
same  relation  to  XX'  and  YY'  as  (d)  to  YY'  and  XX' . 


FUNCTIONS  AND  LIMITS 


13 


The  graphs  of  the  circular  functions  have  the  appearance  of  a curve 
repeated'  over  and  over  as  the  variable  increases  or  decreases.  As  in 
(c)  and  {df if  we  revolve  (e)  and  (/ )~around  XX and  interchange 
XX'  and  YY',  we  shall  have  the  graphs  of  arc  sin  x and  arc  tan.tr 
respectively. 

5.  Limits.  For  the  study  of  the  Calculus  it  is  absolutely 
essential  that  the  student  should  understand  perfectly  the 
fundamental  notion  of  a limit.  He  is  already  familiar  with 
simple  examples  of  limits  from  Geometry,  such  as  the  limit 
of  the  perimeter  of  an  inscribed  regular  polygon  as  the 
number  of  sides  is  indefinitely  increased  is  the  circum- 
ference, and  the  limit  of  the  area  of  the  polygon  is  the 
area  of  the  circle.  These  are  examples  of  variables  ap- 
proaching limits , the  variable  being  in  the  first  case  the 
perimeter,  and  in  the  second  the  area  of  the  regular 
polygon.  The  following  definition  states  the  matter  gen- 
erally. 

Definition.  A variable  is  said  to  approach  a number  A 
as  a limit  when  the  values  of  the  variable  ultimately  differ 


14 


FUNCTIONS  AND  LIMITS 


from  A by  a member  whose  numerical  value  is  less  than 
any  assignable  positive  number. 

If  we  represent  the  values  of  the  variable  by  the  infinite 
sequence 

^l>  ^2’  ^3»  ’"*>  ^ni  fi'n+b  “"> 

then  on  the  scale  (Fig.  i)  the  points  corresponding  to 
av  a2,  a3,  •••,  an,  an+l,  •••,  etc.,  will  ultimately  approach 
nearer  the  point  A than  any  assignable  length,  that  is,  will 
“heap  up”  at  the  point  A.  The  definition  interpreted 


A 


ax 

a2 

|< — h-> 

* — h — *| 

Fig.  4 

geometrically  means,  then,  that  no  length  h (Fig.  4),  however 
small , can  be  laid  off  from  the  point  A,  but  that  points  of 
the  sequence  will  fall  within  the  segment. 

•We  write  Limit  (#„)  = A,  or,  also,  if  we  denote  the  varia- 
ble whose  values  are  av  a2,  etc.,  by  x, 

Limit  (x)>=  A. 

6.  Limiting  Value  of  a Function.  Continuous  Function. 

Consider  the  elementary  function  \ogax  (Fig.  3).  Take 
any  sequence 

^2*  ^3*  * **> 

of  positive  numbers  whose  limit  is  some  positive  number  A. 
For  example,  the  sequence 

i-3.  I-33>  1-333,  — , 

the  limit  of  which  is  f.  Consider  now  the  sequence  of 
numbers 

logo  («i),  loga(tf2)>  log«  Os),  •"> 
and  draw  their  ordinates  in  Fig.  3.  Then  the  student 


FUNCTIONS  AND  LIMITS 


15 


will  see  that  this  last  sequence  has  the  limit  loga(^d); 
that  is, 

When  the  variable  x approaches  a limit  A greater  than 
zero , the  logarithmic  function  logax  approaches  the  limit 

log a A. 

We  express  this  important  fact  by  writing 
Limit  (}ogax)x=A  = lo  gaA. 

The  general  relation  brought  out  by  this  example  is  the 
following : When  the  values  assumed  by  the  variable  x 
approach  * a limiting  value  A,  then  the  corresponding 
values  of  the  function  will  also  approach  a limiting  value ; 
and  if  the  function  is  defined  for  the  value  A,  then  the 
limiting  value  of  the  function  is  its  value  for  x=  A.  Or, 
in  symbols,  if  f (A)  is  a number,  then 

Limit  (f(x)')x=A  =f(A). 

For  example,  since  cos  0=1,  Limit  (cos  x)^  = 1.  The 
property  above  described  is  that  of  continuity;  i.e.  a con- 
tinuous  function  is  such  that 

Limit  f(x)  = /(Limit  x). 

For  the  purposes  of  the  Calculus  it  is  essential  that  a 
function  should  be  continuous.  The  elementary  functions 
of  § 2 possess  this  property. 

7.  Infinity.  If  the  points  on  the  scale  of  Fig.  1 cor- 
responding to  the  sequence  of  values  of  the  variable  x 

* The  variable  x may  approach  the  limit  A in  any  manner  consistent  with  the 
definition  of  the  function.  In  the  above  illustration  the  geometrical  sequence 

1,  1 + 1,  1 -f- 1 + ie,  1 + * + is  + 6s,  etc., 
whose  limit  is  f,  might  also  have  been  taken. 


i6 


FUNCTIONS  AND  LIMITS 


ultimately  advance  to  the  right  without  limit,  we  say,  “ x 
increases  without  limit,”  or  also,  “ x approaches  the  limit 
positive  infinity,”  and  we  write 

Limit  x = + oo. 

If  under  the  same  conditions  the  points  advance  to  the 
left  without  limit,  we  say,  “.r  decreases  without  limit,”  and 
write 

Limit  x = — oo. 


Finally,  if  the  points  advance  both  to  the  right  and  left 
without  limit,  we  write 

4 

Limit  x = oo. 

The  student  should  disabuse  his  mind  of  any  previous 
notions  of  infinity  not  agreeing  with  the  above  definitions. 
The  symbols  + oo,  — oo,  cc,  must  be  used  always  in  the 
sense  above  described. 

8.  Fundamental  Theorems  on  Limits.  The  student  is 
asked  to  accept  the  following  theorems  as  true : 

Given  a number  of  variables  whose  limits  are  known ; 
then 

I.  The  limit  of  an  algebraic  sum  of  any  finite  number 
of  variables  equals  the  same  algebraic  sum  of  their  respective 
limits. 

II.  The  limit  of  the  product  of  any  finite  7iumber  of 
variables  equals  the  product  of  their  respective  limits. 

III.  The  limit  of  a quotie7it  of  two  variables  equals  the 
quotie7it  of  their  respective  limits  when  the  limit  of  the 
denominator  is  7iot  zero. 


1 7 


Fig.  s 


FUNCTIONS  AND  LIMITS 

9.  Two  Important  Limits.  To  prove  * 

Limit  =1. 

L a?  J*=o 

In  Fig.  5 let  x=  arc  A T 
=-.  arc  A S,  the  radius  OA 
being  taken  equal  to  unity. 

Then 

sin  x = MT  = SM, 
tanar  = TQ  = QS. 

Now  ST <nrc  ST <SQ  + QT. 

2 sin  x<  2 x<  2 tan  x\ 
whence,  dividing  through  by  2 sin  x, 

i < x < tan'y  ( — 1 A 

sin  x sin  x \ cos  x) 

Therefore,  taking  reciprocals, 


^ sin  x 

COS  X < < I. 


Now  let  x approach  zero  as  a limit ; then,  since  cos  o=  i, 

id  t 
have 


sin  % 

and  the  value  of lies  between  i and  cos;r,  we  must 


x 


Limit 


Sin  x 
x 


= i. 


10.  Consider  next  the  infinite  sequence 


i,  r +~,  i 


i , i 
> 1 + 


I • 2 


1-2  1-2-3 


not  defined  for  x = o. 


EL.  CALC.  — 2 


i8 


FUNCTIONS  AND  LIMITS 


Representing  the  successive  terms  by  av  a2,  a3,  •••,  we 
have 


— U 

a2  = 1 + Y’ 

aZ  ~ 1 + ~ + 
ai  — 1 + ~ + 
an  — I +~  + 


I 

I • 2’ 
I 

I • 2 
I 

I • 2 


+ 

+ 


1-2-3 
I 

1-2-3 


+ 


= I 
= 2 


= 2.5 

= 2.666  - 

4- , — - — *,  etc. 

\n  — i 


The  numbers  of  this  sequence  continually  increase.  We 
may  show,  however,  that  any  term  is  less  than  3. 

If  r>  3,  then  |r>2r,  and  therefore 

~ 1 

I I I I I 2" 

an<i  1 -\ F H — o d — q + ---H — — 1 H > 

I 2 2i  23  2"  1 I 

I 

2 


since 


1 1 1 

I d 1 n d 5 + 

2 2"  2d 


+ 


I 

2n— 1 


is  a geometrical  progression  and  its  sum  may  be  imme- 
diately written  by  the  usual  formula. 

Hence  a„  < 3 and  taking  n=  1,  2,  3,  etc.  ijifi- 

nitum,  every  term  of  the  sequence  is  seen  to  be  less  than  3. 

c 

CTl  do  Q3 

5 1 S h_ 

Fig.  6 


The  points,  then,  corresponding  to  the  sequence  (Fig.  6) 
must  heap  up  at  some  point  to  the  left  of  3 ; that  is,  the 
sequence  must  have  a limit. 

* The  symbol  \n  — t,  read  " factorial  n — i,”  means  the  product  of  all  integers 
from  1 to  n — i inclusive. 


FUNCTIONS  AND  LIMITS  19 

The  calculation  of  this  limit  to  any  number  of  decimal 
places  is  a matter  of  no  difficulty,  as  the  following  compu- 
tation to  five  decimal  places  will  show. 


Write  down 

1. 000000  ( 

= 1). 

Divide  by 

2 ) 1 .000000  ^ 

4) 

3). 500000  ^ 

II 

4). 166667  ( 

-5) 

5 >.041667  ( 

-b) 

6). 008333  ( 

-b) 

7). 001388  ( 

=|) 

8 >.000198  ^ 

9 >.000025  ^ 

:-D 

10 >.000003  ( 

:=s) 

Adding, 

2.71828 

neglecting  the  figure  in  the  sixth  decimal  place,  of  which 
we  cannot  be  sure.  In  fact,  it  can  be  easily  shown  that 


20 


FUNCTIONS  AND  LIMITS 


2.71828  is  the  limit  of  the  sequence  correct  to  five  deci- 
mal places. 

Writing  the  limit  of  the  sequence  in  the  form  of  an  infi- 
nite series  and  denoting  this  limit  by  e,  we  have 


The  number  e is  called  in  the  Theory  of  Logarithms  the 
Napierian  base  or  natural  base,  and  is  a number  of  prime 
importance  in  mathematics. 

The  expression  for  e in  the  form  of  an  infinite  series 
should  be  remembered  and  also  its  value  to  five  decimal 
places. 

11.  To  prove 


A rigorous  proof  of  this  very  important  limit  is  beyond 
the  scope  of  this  volume.  We  may  perhaps  best  illustrate 
the  meaning  of  the  theorem  by  drawing  the  graph  of  the 
function  for  positive  values  of  z. 


e = 2.71828  .... 


Limit  Tl  + -1  = e. 

L ZJz=OD 


Setting 


and  for  any  value  of  z greater  than  zero  p may  be  approxi- 
mately calculated,  as  for  example  in  the  accompanying 
table,  which  gives  y to  five  decimal  places. 


FUNCTIONS  AND  LIMITS 


21 


z 

y 

.01 

1.04723 

.1 

1.27098 

I. 

2. 

IO 

2.59374 

IOO 

2.70481 

IOOO 

2.71692 

10,000 

2.71815 

100,000 

2.71827 

1,000,000 

2.71828 

The  figure  illustrates  the  theorem  in  showing  that  the 
graph  approaches  the  line  y = e as  z increases  indefinitely. 
When  z diminishes  toward  zero,  y approaches  unity. 


22 


FUNCTIONS  AND  LIMITS 


EXERCISE  2 

[The  graph  of  the  function  considered  must  be  drawn  in  every  case. ] 

1.  Prove  Limit  | - — ~ 3 ^ + 41  _ 2, 

L x-i 

We  have  merely  to  substitute  2 for  x. 

2.  Prove  Limit  [ '*  — — 1 =— 2 a. 

L x+a 

We  cannot  substitute  directly,  for  we  should  get  a meaningless 
2 2 O 

expression.  But  ' = .r  — a,  and  we  may  now  substitute. 

x+a 

3.  Prove  the  following  in  which  a is  any  number  greater  than  zero  ■ 

Limit  1~— 1 = + co  ; Limit  f-1  = oo ; 

Ur 2 J a-o  La-J^ 

Limit  ax  — oo ; Limit  — = o. 

L LjrJ^ 

The  last  three  results  are  often  written 

a a 

- = oo,  a- oo=oo,  — = o, 
o 00 

but  the  student  must  remember  that  such  equations  are  merely  abbrevia- 
tions of  the  preceding. 

~Vx  + h — VjrJ  _ i 


4.  Prove 


Limit 


*[’ 


h= 0 2\/X 


Hint.  Multiply  numerator  and  denominator  by  Vx  + h + Vx. 
5.  Show  that 

Limit  [ tan  '*  1 =i;  Limit  Ttan^rl  = oo  ; 

LsinxJx^  L J*=f 

Limit  Qoge.rJ  = — oo  ; Limit  = o. 


CHAPTER  II 


DIFFERENTIATION 


12.  Increments.  In  order  to  understand  the  manner 
of  variation  of  a function  as  the  variable  varies,  it  is  essen- 
tial to  know  how  great  a change  in  value  occurs  in  the 
function  for  a given  change  in  value  of  the  variable. 
Change  in  value  is  termed  increment ; i.e.  the  increment  of 
the  function  is  the  change  in  value  of  the  function  corre- 
sponding to  a given  change  in  value  or  increment  of  the 
variable. 

The  problem  now  arises  : To  calculate  the  increment  of  a 

given  function. 

Let  fix ) be  defined  for 

all  values  of  x from  x to 
f{x+h)-f{x)  x + h (Fig_  g)  Nqw  for 

x + h the  value  of  the  func- 
tion is  fix  + h),  hence  the 
increment  of  the  function 
fix)  corresponding  to  an 
x~  increme7it  h in  the  variable 


x 

Fig.  8 


x+h 


x is 


fix  + h)  -fix). 


We  shall  represent  the  increment  of  any  variable  by  the 
letter  A (read  “ delta  ”)  prefixed  to  that  variable,  thus 


If  Ax  = h,  then  A fix)  = fix  + ft)  — fix') . 

23 


24 


DIFFERENTIATION 


Rule.  To  find  the  increment  of  a function , ca'culate  the 
new  value  of  the  function  by  replacing  x by  x + h and  sub- 
tract the  old  value  of  the  function  from  the  new  value. 


EXERCISE  3 


1.  Find  Ax 2. 


2.  Find  A 


a)- 


Ax2  = (x  + h)2  — x2  — 2 hx  -f  h2.  Ans. 

A / 1 \ I I —k 

A ( — ) = — — t = — — Ans. 

\x ) x+h  x x(x+h) 


3.  Prove  A \fx=- 


h 


dx-\-h  + Vx 


4.  Find  A lo zx. 


(Ex.  4,  page  22.) 


A log  x — log  (x  -f-  //)  — log  x = log  j =log^i  + Ans. 


5.  Find  Asin.tr. 

A sin  x = sin  (x  + h)  — sin  x = 2 cos  (x+  ^ h ) sin  \ h, 

from  Trigonometry.  Ans. 

6.  Find  Aez.  Ae*  = — e*  = — i).  Ans. 


7.  Find  A cos  2 .r.  A cos  2 x = — 2 sin  (2  .r  + h)  sin  A.  Ans. 

, h a 

8.  Find  AVi  + x , ==  Ans. 

vi+r+/i  + Vi+r 

13.  The  Increment  Quotient.  While  the  increment  of 
a function  as  found  in  the  preceding  article  is  of  impor- 
tance, still  more  essential  in  any  investigation  is  the  rate 
of  change  of  the  function , that  is,  the  change  in  the  function 
per  unit  change  in  the  variable. 

If  we  form  the  quotient 


DIFFERENTIATION 


25 


we  obtain  the  average  rate  of  change  of  the  function  while 
the  variable  changes  from  x X.o>  x h. 

For  example,  the  “ law  of  falling  bodies,” 
given  in  Mechanics,  asserts  that  the  distance  s 
traversed  by  such  a body  falling  freely  from  rest 
in  a vacuum  varies  as  the  square  of  the  time  t , 
that  is, 

s = 16. 1 t2, 

the  constant  16. 1 being  determined  experimen- 
tally when  is  measured  in  feet  and  t in  seconds. 
Therefore  As  = 16.1  (t  + Jif  — 16. 1 t2, 

or  — = 16.1(2  t + h),  since  At  = h. 

At  v ’ 

For  example,  the  average  velocity  throughout 

/\  c 

the  third  second  is  given  by  setting  in  — , t — 2, 
h—i,  and  is  80.5  feet  per  second. 

EXAMPLES 

1.  From  Physics  we  learn  that  for  a perfect  gas  at  constant  tem- 
perature the  product  of  the  pressure  p and  volume  v is  constant,  or 

, . , c , , \t>  c 

= a constant  c,  i.e.  t>  = - : show  that  = , ; — 

v tsv  v 2 + vtSv 

2.  Show  from  Ohm’s  law,  viz.  current  strength  C equals  electro- 
motive force  E divided  by  the  resistance  R , that  for  constant  R the 
change  of  current  strength  per  unit  change  of  electromotive  force  is 
constantly  equal  to  1 ~ R. 

14.  Derivative  of  a Function.  In  the  illustration  taken 
from  the  law  of  falling  bodies  given  in  § 13,  let  us  propose 
to  ourselves  to  find  the  velocity  at  the  end  of  two  seconds. 
Making  t=  2,  we  have 


t= 3 

Fig.  9 


t=  0 


f= 3 


26 


DIFFERENTIATION 


which  gives  us  the  average  velocity  throughout  any  time 
h after  two  seconds  of  falling.  Our  notion  of  velocity 
shows  us,  however,  that  by  the  velocity  at  the  end  of  two 
seconds  we  do  not  mean  the  average  velocity  during  one 
second  after  that  moment,  or  even  during  or  ^oVo 
of  a second  after  that  moment,  but,  in  fact,  we  mean  the 
limit  of  the  average  velocity  when  h diminishes  toward 
zero ; that  is,  the  velocity  at  the  end  of  two  seconds  is 
64.4  feet  per  second.  Thus,  even  the  everyday  notion 
of  velocity  involves  mathematically  the  notion  of  a limit, 
or,  in  our  notacion, 


Thus,  after  t seconds  have  elapsed,  the  velocity  of  a 
falling  body  is  32.2  t feet  per  second. 

Again,  let  it  be  required  to  find  the  slope  of  the  tangent 
at  any  point  P of  a plane  curve  whose  equation  is  given 
in  rectangular  coordinates  x and  y. 


The  tangent  at  P is  constructed  as  follows  (Fig.  10): 
Through  P and  any  point  P'  on  the  curve  near  P draw 
the  secant  AB.  Let  the  point  P'  move  along  the  curve 


Velocity  = Limit 


DIFFERENTIATION 


V 


toward  P,  the  secant  AB  meanwhile  turning  around  P. 
Then  when  P'  coincides  with  P,  the  secant  AB  becomes 
the  tangent  TP. 

Now  if  P is  (x,  y)  and  P ' ( x + Ax,  y + Ay),  the  slope  of 
AB  is 


tan  6 = 


SP'  = Ay 
PS  Ax 


As  P’  approaches  P as  above  described,  Ax  will  ap- 
proach zero  as  a limit,  while  6 approaches  the  angle  PTO 
or  7 ; hence,  at  the  limit, 


tan  -y  = Limit  j ^ = slope  of  tangent  at  any  point  P. 

For  example,  the  slope  of  the  tangent  at  any  point  of 
the  parabola  y = x 2 + 3 is  2 x. 

Law  of  Linear  Expansion.  If  l0  is  the  length  of  a rod  at  o°  Centi- 
grade, and  l the  length  at  t°  on  the  same  scale,  then  experiment  estab- 
lishes the  law  of  expansion 

/ = /q  - f-  at  bE, 

a and  b being  constants.  The  coefficient  of  linear  expansion  at  any  tem- 
perature t is  the  increase  in  length  per  unit  change  in  temperature,  i.e. 


coefficient  of  expansion  = Limit  ( — ’] 

\ A/  / £,=0 

We  easily  find,  then,  that 

coefficient  of  expansion  = a + 2 bt, 
and  .\  a — the  coefficient  of  expansion  at  o°. 


Specific  Heat  of  a Substance.  The  specific  heat  of  any  substance  is 
the  quantity  of  heat  necessary  to  raise  a unit  mass  of  the  substance 
one  degree  in  temperature.  If  Q is  the  measure  of  the  quantity  of  heat 
in  unit  mass,  and  t the  corresponding  temperature,  then  by  definition, 


specific  heat  — Limit 


28 


DIFFERENTIATION 


These  examples  show  that  we  obtain  an  important  new 
function  of  the  variable  if  we  can  find  the  limit  of  the 
Increment  Quotient  when  the  increment  of  the  variable 
approaches  zero.  This  function  is  called  the  derivative 
of  the  function. 

Definition.  The  derivative  of  a function  is  the  limit  of  the 
quotient  of  the  increment  of  the  function  and  the  increment  of  the 
variable  when  the  latter  increment  approaches  the  limit  zero. 

The  step  of  finding  the  limit  of  when  Ax  ap- 

Ax 

proaches  o is  indicated  by  changing  the  A’s  to  ordinary 
d’s,  so  that  df-xl  = Limit  ^ , or  also,  if 

dX  \ AX  / Aar=0 


Ax  = h,  = Limit  [ 


+h)-f(x)l 


h 


J/i=0 


The  symbol  c^S:x.l  is  read,  “ derivative  of  f(x)  with 
dx 

respect  to  x.”  This,  being  a new  function  of  x , is  often 
written  f (x),  so  that  also, 


4f(*) 

dx 


=/'(*> 


Thus  in  the  illustrations  given, 

velocity  = — , 

J dt 

i.e.  velocity  is  the  derivative  of  the  space  traversed  in  the 
time  t with  respect  to  the  time. 


Slope  of  tangent  = 

dx 


DIFFERENTIATION 


29 


i.e.  equals  the  derivative  of  the  ordinate  of  the  point  with 
respect  to  its  abscissa. 

Coefficient  of  linear  expansion  = — , 

dt 

or  the  derivative  of  the  length  with  respect  to  the  temperature. 

Specific  heat  = 

dt 

that  is,  equals  the  derivative  of  the  quantity  of  heat  in  unit  mass  with 
respect  to  the  temperature. 

Many  more  illustrations  of  physical  magnitudes  might  be  given  which 
take  the  form  of  a derivative. 

We  call  — the  sign  of  differentiation , so  that  the  pre- 
dx 

fixing  of  — to  any  function  of  x means  that  the  following 
dx 

process  is  to  be  carried  through  : 

General  Rule  of  Differentiation.  1°.  Calculate  the 
quotient  of  the  increment  of  the  function  and  the  increment  of  the 
variable  (i.e.  the  increment  quotient). 

2°.  Find  the  limit  of  this  quotient  when  the  increment  of  the 
variable  approaches  the  limit  zero.* 

It  must  be  emphasized  here  that  the  characteristic  thing 
in  differentiation  is  finding  the  limit  of  a quotient.  From 
the  standpoint  of  the  Differential  Calculus  a function  is  of 
no  interest  if  the  limit  mentioned  does  not  exist.  Func- 
tions possessing  derivatives  are  said  to  be  differentiable , 
and  it  is  of  prime  importance  to  show,  for  example,  that 
the  elementary  functions  of  § 2 are  differentiable. 


* The  student  must  notice  that  the  limit  of  the  increment  quotient  cannot  be 
found  by  Theorem  III,  § 8,  since  the  limit  of  the  denominator  is  zero. 


30 


DIFFERENTIATION 


15.  Differentiation  of  the  Elementary  Functions 
xm,  sin  x,  loga  x. 

(a)  To  prove  —x m = rnxm~1. 
dx 

Now  A(xm)  = (x  + Ji)m  — xm  if  Ax  = h. 

But  (x  + li)m  = xm  + inxm~1h  H \- hm, 

the  terms  not  written  containing  powers  of  h. 

A(xm)  = mxm~lh  1-  hm ; 

= vixm~x  -} f Am~\ 

Ax 

where  again  the  terms  not  written  contain  powers  of  h. 
Putting  h — o,  we  find 


(0 


d_ 

dx 


(xm)  = mx‘ 


.771  — 1 


(b)  To  prove — sin  ;tr  = cos  ;r. 
dx 

Since  A sin  x — sin  (x  + A)  — sin  x 

= 2 cos  ( x + J Ji)  sin  \h  (§  12,  Ex.  5), 

we  find 

A sin  x _ 2 cos  (x  + ^ h)  sin  \ h 
Ax  h 

= cos  (x  + j h)  • — 2— 

2 

But  Limit  (— y-f— ^ =I  (§9). 

\ \ ll  Jh= 0 

and  Limit  (cos  (x  + ^/i))h==0  = cos 

(since  cos  x is  continuous,  § 6), 
so  that  we  may  apply  the  theorem  II,  § 8,  and  we  have 


(2) 


- sin  x = cos  x. 
dx 


DIFFERENTIATION 


3* 


00  To  prove  loga  x = loga  e ~ 
From  § 12,  Ex.  4,  we  have 


A 10ga  X 

Ax 


loga^i 

IT 


=-1Qg* 


k\h 

I _l — I* 

Xj 


since  the  introduction  of  the  exponent  — is,  by  the  princi- 
ples of  logarithms,  equivalent  to  multiplying  the  logarithm 

itself  by  that  exponent. 

/ /A* 

Now  ( 1 + - r is  the  expression  of  § 1 1 if  we  write  in  that 


expression  z = — 

h 

Also 


Limit 


A=0 


hence  Limit 

rr.Af 

— Limit 

[7.+±Yl 

_V  x)  _ 

h= 0 

Lv  z)  J 

= e. 


Limit 


bcr 

a 


1 H — )ft 

Xj 


h= 0 


= loga  e. 


and  we  have 
(3) 


(since  logx  is  a continuous  function), 

4-  log« x — lo&« e 

ax  x 

Formula  (3)  becomes  most  simple  when  a — e,  for  then 

d , 1 

— logex  = ~ 
dx  x 

Logarithms  to  the  base  e are  called  natural  logarithms  or  Napierian 
logarithms  (§  10),  and  the  factor  loga  e in  (3)  is  called  the  modulus  of 
the  system  whose  base  is  i.e.  the  member  by  which  natural  logarithms 
enust  be  multiplied  in  order  to  obtain  logarithms  to  any  given  base  a. 

We  write 


32 


DIFFERENTIATION 


For  example,  the  modulus  of  the  common  system  of  base  io  is  log;o  e, 
and 

logio  e = 0.43429 

to  five  decimal  places. 

If  in  (1),  (2),  and  (3)  we  write  u for  jr,  we  have 

(4)  um  = mum~1 ; sin  u = cos  v ; ~~  loga  u = loga  e 1 • 

7 du  du  du  u 


EXERCISE  4 

1.  Differentiate  with  respect  to  x. 

( a ) x2  + 3 ax+  b. 

(, b ) — — . 

(r)  Vir  (cf.  Ex.  4,  p.  22). 


2.  Prove — cos  .r  = — sin  x. 

dx 

3.  Prove  — Vi  + x= 1 

dx  2 V 1 + x 


4.  Prove 


d fa\_  a 
dv\v 


we  found  (§  14) 
or 

Prove 
What  does 


ds 

dt 


= 32.2 1, 


velocity  = v = 32.2 1. 


dv  „ „ 

— = 32.2. 
dt  J 

— = Limit 
dt 


Ans.  2 3 a. 


A us. 


(x+ b)'2 


Ans. 


2Vx 


5.  Prove  — {Cu')  = C,  if  C is  any  constant. 

du 

6.  From  the  law  of  falling  bodies 

s = 16. 1 12, 


AzA 

A/  y ^^=0 


represent  in  Mechanics  ? 


Acceleration.  Ans. 


DIFFERENTIATION 


33 


7.  Find  the  velocity  and  acceleration  of  the  motion  defined  by 

(1)  s = at  + \gt2\  Ans.  V=a+gt ; accel.  =g. 

(2)  s = at  — \gt2-,  Ans.  V = a — gt ; accel.  = —g. 


8.  Find  the  slope  of  the  tangent  to  y = 6 x — x2  at  the  origin. 

Ans.  6. 


16.  Certain  General  Rules.  We  prove  in  this  section 
several  important  rules  for  differentiation  of  a general 
character. 

Let  the  variables  u,  v,  zv,  etc.,  be  functions  of  the  vari- 
able ;r. 

I.  To  differentiate  any  algebraic  sum  of  these  variables. 


For  example,  to  find  — (u  + v — w). 

dx 

Now 


A(u  + v — w)  = u + A u + v + Av  — (zv  + Azv)  — {tt  + v — zv), 

= An  + Av  — Azv, 

A(zt  + v — zv)  _ Ate  Av  Azv 

Ax  Ax  Ax  Ax 


Since  Limit 


Au 

= — , Limit 

Av 

_Ax_ 

A3'=o  dx 

_Ax_ 

_ dv 
^^=0  dx 


Limit 


Azv 

A x 


_ dzv 
— 0 dx 


we  may  apply  I,  § 8,  and  we  have 


(5) 


d , , x du  , dv 

— (u  + v-zv)=  — + — - 
dx  dx  dx 


dzv 

dx 


II.  To  differentiate  a product. 

For  example,  to  get  ~ (uv). 

dx 


EL.  CALC.  — 3 


34 


DIFFERENTIATION 


Now  A {uv)  — {u  + A u)(v  + Av)  — uv, 

= u Av  + v A u + A u Av, 

A (uv)  Av  . An  . A Av 

••  a — + + 

Ax  Ax  Ax  Ax 

c.  T.  ..  ["AzTI  dv  T.  ..  fAz/l  da 
Since  Limit  - — = — , Limit  — = - — -, 


jA^=0 


dx 


\_Ax\ 

Limit  [Az*]ax=0=  o, 
we  may  apply  I and  II,  § 8,  and  obtain 

(6) 


_AxJ 


Iax=0 


dx 


d , \ dv  , dn 

— {iiv)=u  — + v~ 
dx  dx  dx 


To  find  — (uvw),  consider  nvw  as  made  up  of  the  two 
dx 

factors  uv  and  w ; then,  by  (6), 

d , \ dw  . d(uv) 

— (uv  ■ w)  = uv b w — — % 

dx  dx  dx 


or  by  (6)  again, 

(7) 


dw  , dv  , du 

= uv  — — b wu  ——  + wv  — — 
dx  dx  dx 


III.  To  differentiate  a quotient. 


Since 


a r-  = 


u + An 


v Au — u Av 


v + Av  v v2  + v Av  ’ 


Au  Av 

v u - — 

A fu\_  Ax  Ax 

A x\vj  v2  + vAv 


Then  since  Limit  [_v2  + vAv~\^x=S)=  v2,  we  may  apply  I, 
II,  III,  § 8,  and  have 

du  _ dv 

(8)  d fu\  _ dx  dx 

dx  \vj  v2 


DIFFERENTIATION 


35 


From  (6),  (7),  and  (8)  we  have  the  rules  : 

I.  The  derivative  of  an  algebraic  sum  of  any  number  of  vari- 
ables is  equal  to  the  same  algebraic  sum  of  the  derivatives  of  the 
variables. 

II.  The  derivative  of  a product  of  any  number  of  variables  is 
equal  to  the  sum  of  all  the  products  formed  by  multiplying  the 
derivative  of  each  variable  by  ail  the  remaining  variables. 

ITT.  The  derivative  of  a quotient  equals  the  denominator  times 
the  derivative  of  the  numerator  minus  the  numerator  times  the 
derivative  of  the  denominator,  all  divided  by  the  square  of  the 
denominator. 

To  these  we  may  add  the  following: 

IV.  The  derivative  of  a constant  is  zero. 

V.  The  derivative  of  a constant  times  a variable  equals  the 
constant  times  the  derivative  of  the  variable. 


Rule  V comes  from  (6)  and  IV,  if  we  place  u equal  to  a 
constant. 

EXAMPLES 

1.  Workout  — (1  + x2)  (1  — 2x2). 

dx 

Rule  II  is  first  applied,  and  we  get 

— (1  + X2)(l  -2r2)  = (l  + X2)  — (I  - 2X2)  + (l  - 2X2)  — (I  + X2). 
dx  dx  dx 

By  Rule  I,  £ 0 - »**)  = £ (0  ~ £(***>> 


d_ 

dx 


(i  + x2) 


Since  by  V, 


— (2  x2)  = 2 — x2,  and  from  (4)  § 15, 
dx  dx 


= 2r, 


we  have  finally, 


d_ 

dx 


(1  -f  x2)(l  — 2X2)  = (1  +x 2)  • — 4 x+  (1  — 2 X2)  -2X=  -2r(l+  4X2). 


36 


DIFFERENTIATION 


2.  Work  out 


d ! sin  x \ 
dx\\oge  x) 


Rule  III  we  use  first  and  find 


d / sinr\ 
dx  Vloge  x) 


log,  x — sin  x — sin  x — loge  x 
dx  dx 


0°geX)2 


By  (4)  § 1 5j 


d • d , i 

— sin  x — cos  x,  — lo£,  x = - • 
dx  ’ dx  6 * 


/ sin  .r  \ ;r  cos  .r  log,  x — sin  x 

dx  \ loge  x)  x (loge  x)  2 


EXERCISE  5 


Prove  the  following  differentiations  : 

1 d \ -j  d fsinx\  _xcosx-smx 

dx^1-^-1-2*-  3-  dx[-r)- — ^ — 

2 d ( x \ - I — x2  4 d / 1 + .r2\  _ 4J 

<faAi+^2/  (i+ur2)2  */rVi  — x2J  (i  — x2)1 

g d / xm  \ _ xm~1(m  + xn  (m  — «)) 
rthr\i+.rn/  (i+x’*)2 


6. 

7. 

8. 


— (urm  log  x)  = x-m  l(i  + in  log  x) . 
dx 

d_  ( 2 x \ _ 2 ( i + x2) 
dx\i  — x2)  (i  — x2)2 

<L  (*?)  = !Lxn-i.  9 . 

dx\a  / a dx  \jrn 


— na 

xn+l 


10. 


— xm  (i  — x)n 
dx  K ’ 


— xm- 1 ( t _ x)"-1  (in  — (m  + n)x) . 


Special  attention  should  be  given  to  the  following : 

11.  Find  — tanw.  Since  tan  u = sm— , 

du  cos  u 


we  have 
Applying 


— tan  u 
du 


d ( sin  u 
du  Vcos  u 


III  (4),  § 15,  and  Example  s,  § 15,  we  find 


— tana  = sec2«. 
du 


DIFFERENTIATION 


37 


Prove 


12.  ■ — cot  u — — cosec2  u. 
du 


^ Put 


sec  u — 


cos  u 


I 


:•) 


14.  — CSC  U = — CSC  21  cot  u. 
du 


17.  We  come  now  to  two  most  important  rules. 
Differentiation  of  Inverse  Functions.  Suppose  y is  a 
function  of  i.e.  in  symbols 


Then  it  is  usually  possible  inversely  to  calculate  when 
values  are  assumed  for  y,  i.e.  we  may  choose  y for  the 
independent  variable  instead  of  ;r,  so  that  by  solving  (9) 
for  x we  obtain 


Then  fix)  and  <£(j y)  are  called  inverse  functions. 

Example.  If  y — ax,  then  x = lo gay ; that  is,  a1  and  loga_y  are 
inverse  functions. 

Let  now  and  Ay  be  corresponding  increments  of 

and  y,  so  that  Ax  and  Ay  vanish  together,  since  we  are 

dealing  here  with  continuous  functions.  Then  the  incre- 

Ay  Ax  , 

ment  quotient  is  or  , according  as  .r  or  y is  taken 

for  the  independent  variable. 


(9) 


y =/(•*)• 


(10) 


x = <Ky)- 


Now  by  multiplication,  ^ 


hence  Limit 


Limit 


38 


DIFFERENTIATION 


by  II,  § 8,  since,  as  above  emphasized,  A y and  Ax  vanish 
together.  We  have,  therefore,  in  the  derivative  notation, 


(ii) 


dy  dx  , . 

~dx ' ~dy  = * ’ or  solving,  ' 

* 

dy  ~ dy 
dx 


YI.  If  y is  a function  of  x,  and  inversely  x a function  of  y,  then 
the  derivative  of  x with  respect  to  y equals  the  reciprocal  of  the 
derivative  of  y with  respect  to  x. 

Differentiation  of  a Function  of  a Function.  We  have 
seen  by  (4)  how  to  differentiate  with  respect  to  ;r  the  ele- 
mentary function  sin.r.  Suppose  we  wish  to  find 

~r~  sin  ( 1 + x2), 


for  which  the  rule  (4)  does  not  suffice.  We  then  introduce 
the  variable  u = 1 + x2,  and  setting  y = sin  ( 1 + x2)  = sin  u, 
we  have  before  us  the  relations 


(12) 


y = sin  u,  u — 1 + x2, 


and  we  say  y is  a function  of  x through  u,  i.e.  y is  a function 
of  a function. 


Now,  if  Ay,  A u,  and  Ax  are  corresponding  increments  of 
y,  u,  and  ^r,  then  forming  the  increment  quotients 


we  have,  by  multiplication, 


Au’  Ax 


Ay  Au  Ay 
Au  Ax  Ax 


* The  student  will  not  fail  to  notice  that  in  (n)  the  familiar  property  of  a 

fraction,  %=  i -e-  - is  suggested.  But  he  must  not  forget  that  ^ is  not  a fraction, 
b a dx 

\y 

but  merely  the  symbol  for  the  limiting  value  of  the  fraction  -f— 


DIFFERENTIATION 


39 


But  the  increments  Ay,  A u,  and  Ax  vanish  together,  so 
that,  by  II,  § 8, 


Limit 


(f) 


Limit 


Am=0 


(—)  ■ 
V^-f/Ax=0 


Limit 


©,  ■ 


' Ax=0 


or  (14) 


dy  _ dy  du 
dx  ~ du  dx * 


VII.  If  y is  a function  of  x through  u,  then  the  derivative  of  y 
with  respect  to  x equals  the  product  of  the  derivatives  of  y with 
respect  to  u and  of  u with  respect  to  x. 

Thus  in  (12),  since  — sinzz  = coszz,  — (1  + x2)  — 2 x,  we  find 
d du  dx 

— sin  Ci  + x2)=  cos u • 2x—2x cos  C 1 + x2) . 
dxK 

EXERCISE  6 


1.  Show  that  the  geometrical  significance  of  (11)  is  that  the  tangent 
makes  complementary  angles  with  XX'  and  YY' . 

2.  If  a material  point  P,  whose  rectangular  coordinates  are  x and  y, 
move  in  a plane,  then  x and  / are  functions  of  the  time  t.  Now  the 


horizontal  component  vx  (see  Fig.  11)  of  the  velocity  v is  the  velocity 
along  OX  of  the  projection  M of  P,  and  is  therefore  the  time  rate  of 


40 


DIFFERENTIATION 


change  of  x,  or  vY  = . In  the  same  manner,  the  vertical  component 

, dt 

v2  equals  and  since 

v = Vvj2  + v 22, 


we  have 


For  the  direction  of  the  velocity,  tany  = — , or 


dy  dx 
tan  y = -4  -f-  — . 
dt  dt 


3.  Prove  that  the  equations  x — a cos t,  y = a sin/  define  uniform 
motion  in  the  circle  x'2  + y'2  = a2. 

4.  If  the  coordinates  (x,  y)  of  a point  on  a curve  are  functions  of  a 
variable  0,  show  that 

/ T r \ ^ d_f. 

' dx  dd  dO 

(Use  (14)  and  then  (11).) 


7 _ 

5.  To  find  — (xq)  when  q is  any  positive  integer;  i.e.  to  differ- 
dx 

entiate  any  root  of  x. 

1 

Put  u = xi,  then  x = ui\  hence,  by  (4),  § 15, 


and  by  (11) 


But 


dx  „ , 
— = qun~\ 
du 


du  _ 1 

dx  qui~x 

. *=!  i-l 

7lq~l  — X q — X q. 


du 

dx 


or 


d , 1 l-i 

— (ar?)  = - x* 
dx  q 


i.e.  the  same  rule  holds  for  roots  and  powers  of  the  independent 
variable. 


DIFFERENTIATION 


41 


6.  From  (4),  § 15,  and  VII,  we  have 


(16) 


ci  / d / du  mi  dii 

— ( um ) — — - (um)  — = mu — 

dx  du  dx  dx 

d ■ d , ■ -.du  du 

— sin  it  = — (sin  u)  — = cos  u — 

dx  du  dx  dx 

du 

d , d ,,  .du  . dx 

TxXo^u  = dii^7l)Tx=Xo^e~ 


7.  To  find  — (xq). 

dx 


Letting  u = xq  (Example  3),  we  have 


up  = xi. 


— (.r?)  = — jip  — puv~x  — ( Example  6), 
dx  dx  dx 

p-i  , 1 

= PX  q 

, — 1 1 / ^-1 

= fix  1 . - XI  =£--Xl  . 

q q 

Hence  the  rule  of  (4),  § 15,  for  powers  holds  for  any  commensurable 
exponent. 

du 

mm,  d dx 

8.  To  prove  , - arc  sin  u — — • 

dx  Vi  - u* 

Placing  / = arc  sin  u,  we  have  inversely, 

u — sin/. 


But 

Hence 


~ = cos/,  and  by  (15),  ^ . 

dy  du  cos  / 

cos/  = V 1 — sin2/  = Vi  — u2. 

dy  = 1 

du  Vi  - id 


du 

dy  d dx 

dx  dx  Vi  — id 


and  by  (16), 


42 


DIFFERENTIATION 


du 

9.  Prove  ~ arc  tan  u = . 

dx  i + u2 


(Remember  sec2j  = i + tan2_y.) 

du 


10.  Prove  arc  sec  u — 
dx 


dx 


uVu2  — 1 

11.  Prove  — au  = au  loge  a — • 
dx  dx 

Putting  y = au,  we  have  inversely, 

u = \ogay. 


% = log°e],  (§!5,  (4))- 


dy  _ y _ <z“ 
du  log0  e logo  e 
dy  _ d _ 
dx  dx 


«u  loge  a —■ 
8 dx 


In  particular, 


d M _ afo 


Example  1 1 shows  that  the  exponential  function  ex  possesses  the 
remarkable  property  of  being  its  own  derivative , for 

d _ „dx 
— ex  — c1-  — = cx, 
dx  dx 


In  general,  if  a is  any  constant,  then 


(0 


— eax  = aeax,  since  — (ax)  = a ; 
dx  dx 


that  is,  the  derivative  of  the  function  eax  is  proportional  to  (i.e.  a times) 
the  function  itself.  For  a reason  now  to  be  explained,  the  function  e** 
is  said  to  follow  the  Compound  Interest  Law. 

If  P dollars  be  drawing  compound  interest  at  r per  cent,  then  in  the 
. . . y 

time  A t the  interest  is P\t,  and  hence  the  change  in  P or  A P is 

, 100  ° 

given  by 

A M'=  — P\t.  or  — P- 

100  A t 100 


* From  the  principle  in  logarithms,  logca  - 


loga  C 


DIFFERENTIATION 


43 


Now  suppose  the  interest  to  be  added  on  continuously , and  not  after 
finite  intervals  of  time  A/,  i.e.  we  make  A t approach  the  limit  zero,  and 
conceive  of  P as  increasing  continuously ; then 

dP  r p 

“j7  — r' 

dt  ioo 


so  that  a sum  of  money  accumulating  continuously  at  compound  inter- 
est has  precisely  the  property  above  enunciated  in  (i),  viz.  its  derivative 
is  proportional  to  the  sum  itself. 

18.  From  the  examples  in  Exercises  5 and  6 and  the 
Rule  VII,  we  deduce  the  following  fundamental  formulae 
for  differentiation : 


VIII.  u™  = mu™-- 1 (m  any  commensurable  number) . 

(toe  (tOC 

du 

-jr~v  ( ~L  | ^ doc 

IK.  —logaU=\ogae—. 

X.  — au  = a«  loge  a~^  (a  any  positive  constant). 

(toe  (toe 

XI.  d sin  u = cos  u — • 
dx  dx 

XII.  ~ cos  m = - sin  m — . 
dx  dx 

XIII.  tan  u = secs  u 

dx  dx 

XIV.  -^-C0tM  = -CSC2  M^“. 

dx  dx 

XV.  ~ sec  u = sec  u tan  u — • 
dx  dx 

XVI.  esc  u — - esc  u cot  u — • 
dx  dx 

du 

XVII.  d arc  sin  u = -. 

dx  Vl  - m2 


XVIII. 


— — arc  cos  u = - 
dx 


du 

dx 

Vl  - m2 


44 


DIFFERENTIATION 


du 

XIX.  ~ arc  tan  u = ^ 

dx  1 + m- 


d/u 

XX.  -^-arccotw  = -r^7. 
dx  \ + u* 


XXI. 


— — arc  sec  w = 
dx 


du 

dx 

mVm2  — 1 


XXII. 


-----  arc  esc  n = — 
dx 


du 

dx 

u VuZ  - 1 


The  formulae  and  rules  I-XXII  the  student  must  memo- 
rize. With  their  aid  differentiation  of  the  commoner  func- 
tions is  made  rapid  and  easy,  but  perfect  familiarity  with 
them  is  indispensable. 

To  show  the  application  of  the  rules  three  examples  are 
now  given  : 


1.  Find  1 r 


V i -f-  x2 


By  HI,  4-(- 


d f I — x 


dx\y/ 1 -f  X- 


V i + x2~(*  -4)-(i  -x)  — -v/iT 


dx 


i + x* 


By  I and  IV,  — (i-x)  = -i; 
dx 

from  VIII,  —(i  + ;r2)*  = f(i  +x2)~i-^-(i  + x2), 
dx  2 dx 


and  since 


— ( i + x2)  = 2 x,  we  have 
dx 


d ( i — x \ Vi  + x2  ■ — i —(i  — x)  (i  + x2)~-  • x 

(i  + x2) 


dx[VT+x> 

To  simplify,  multiply  numerator  and  denominator  by 


(I  + x2)2. 


DIFFERENTIATION 


45 


Then,  since  (i  + x2)0  — i,  we  have,  reducing, 


d / i - 
dxy  VY 


i + * 


+ xJ  (i  + x2y 


2.  Find 


dx  ' i 


cos  .r 


+ cos  x 


For  convenience,  set  / = loge"^j — — 


cos 


Since  log, 
then  by  I and  V, 


4 


— COS  XI,.  . I . , . 

= - log  ( I — cos  x)  — log  ( I + COS  X), 

+ cosr  2 sV  ' 2 45  v ' 


S = i - c°s*)-i£j°g.(i+cos*). 


Applying  IX,  we  have 


— (i  — cosx)  — (i  + cosr) 
dy  i dx  I dx 

dx  ~~  2 i — cos  .r  21+  cos  .r 


, and  by  XII, 


dy  i [ sin  x | sin  x \ _ sin  x _ i 
dx  2 \ i — cos  x'  i + cos  x)  i — cos'2  x sin  x 

...  A loge  J1  T_cgi£=_I_. 
dx  ' x + cos  .r  sin  x 


3.  Find  ^ arc  tan 

Setting  the  function  equal  to  y.  we  have,  by  XIX, 

die 9 — ee  + e~e 

dy  dQ\  2 / 2 , 

dd~  (e*-e-o\2~  4 + e2e-2+e-2o^Y  } 

1 + [ 2 / i 

2 

_ ^ + 


46 


DIFFERENTIATION 


EXERCISE  7 

Prove  the  following  differentiations  : 

i d , o , , \ r~ 5 7 xi  — 2 jr2  — I 

1.  — (;rz  + i)  Vxa  — x = . 

dx  2(x3  — _r)£ 

2.  A-Jj  + 1 

’ r * l 


dx'1!  - x (j  _ jr)  Vi  - jr2 

3.  At.  x 


dx\  \J  i . 


i)=7t 


x2l  (l-a:2)2 


4 / 3X3  + 2 \_ 2 

^ \ar(j.-3+  i)t  / jr2(ar3+i)l 

5.  — (i  — 2 x + 3 ar2  — 4_r3)(i  + a;)2=  — 20  x3(l  + x ). 
dx 

6.  ~(i  — 3 .r2  + 6 ^4)(i  + x2)3  = 60  .r5(i  + jr2)2. 
dx 

7-  -^(5i2+2i)=  2(4- + i)5l2+2zl°gc5- 
ax 


8.  — xnax  = xn~lax(n  4-  x\o ge  a ). 
dx 


x 

( 1 - T)2’ 


9 dLrxio§&x+  j (I  _ xy\  = Jog 

« jr  Li— or  J(i  — 

10.  _ 2£?  + 6*  6 \ ajr3^. 

V a a2  a3) 

11.  — logeO*  + e-*)  = e*  ~ e—. 

dx  6 v 7 e*  + e~* 

12.  loge(i  + Vx))  = l——.. 

dx  2(1  + Vx) 

13.  A tan2  5 0 = io  tan  5 d sec2  5 d. 
dd 

14.  4;  sin3  & cos  $ = sin2  $ (3  cos2  $ — sin2  0) . 
dd 

15.  A log  sec  6 = tan  6. 
dd 


DIFFERENTIATION 


47 


16.  (tan2  6 — log  sec2  6)  — 2 tan3  9. 
da 

17.  sin  n6  sinn  6 = n sin”-1 6 sin  {n  + I )0. 
da 

18.  — arc  sin (3^-4 Xs)  = 3 

^ VI  

to  d x a 

19.  — arc  sec  — = — . 

^ & xVx2  — a!2 


20.  — arc  esc 

dx  2 ;r2  — 1 


21. 


a 

— ■ arc  sin 
dx 


1 — x 2 
1 + x2 


2 

I + X2' 


22. 


— arc  tan 
dx 


x + a 
1 — ax 


1 

1 + x2' 


23. 


24. 


25. 


d e?  — e~x  — 2 

— arc  cos — = — 

dx  ex  + e~x  e?  + e~ 


d I 2 — 1 

— arc  sec  \ = - — - 

dx  ' 1 T x 2 \J  1 ~ 


( arc  cot  - + loge  \l- — - ) = 
dx  \ x 'x  + a / 


2 ax2 
x 4 — a 4' 


19.  Differentiation  of  Implicit  Functions.  If  an  analytic 
relation  is  given  between  two  variables  not  solved  for  either 
variable  in  terms  of  the  other,  then  either  variable  is  said 
to  be  an  iviplicit  function  of  the  other. 

For  example,  in  x1  — y1  + 9 = o either  x or  y is  an  im- 
plicit function  of  the  other  variable. 

In  such  a case  either  variable  may  be  chosen  for  the 
independent  variable,  and  if  we  can  solve  explicitly  for  the 
other  (as  in  the  above  example  for  y,  giving  y = Vx2  — 9), 
then  we  can  differentiate  as  before.  But  it  is  generally 
better  not  to  solve  the  equation,  but  to  differentiate  the 
given  relation  as  it  stands. 


48 


DIFFERENTIATION 


Thus,  to  find  ~ from 
ax 


x2  — 3 xy  + 2 y2 


Then 


dy\ 


dy 


2x-3[y  + x^j+4y±  = o, 


dx 


and 


dy  -2X~  2>y 

dx  3x-4y' 


To  justify  this  process  is  beyond  the  limits  of  this  text- 
book. One  thing  is  to  be  noted,  namely,  that  only  those 
values  of  the  variables  which  satisfy  the  original  relation 
can  be  substituted  in  the  derivative. 


exercise  8 


dy 


Find  -d-  from  the  following  equations  : 

1.  y2  — 2 xy  = a2.  Ans. 


dy  y 

dx  y — x 


x 2 V2 

2.  — j + — = I. 

a2  b2 


Ans. 


dy 

dx 


b2x 


3.  ax2  + 2 bxy  + cy2  + 2 fx  + 2 ay  + h = o. 

Ans.  y = _?£jj l±f. 
dx  bx  + cy  + g 


4.  xs  + y3  — 3 axy  = o. 

5.  x$  +yi  = a%. 


. dy  x-  — ay 

Ans.  -y-  = 7, 

dx  y-  — ax 

a dy  yi 

Am-  s=— r 


X* 


6.  Given  r = a( i — cos  6) ; show  = a sin  6- 

dv 


7.  Given  r2  = a2  cos  2 6 ; show 


dr 

Jd 


a-  sin  2 i 


DIFFERENTIATION 


49 


20.  Derivatives  of  Higher  Orders.  Since  the  derivative 
of  a function  of  a variable  x with  respect  to  x is  also  in 
general  a function  of  jr,  we  may  differentiate  the  derivative 
itself,  that  is,  carry  out  the  operation, 


d_ 

dx 


This  double  operation  is  indicated  by  the  more  compact 
notation, 

d2  r,  x 


and  this  new  function  is  called  the  second  derivative. 
the  same  way, 


d d2  r,  s d 3 ,,  \ 


In 


is  the  third  derivative,  and  in  general, 


dn 

dx“ 


fix) 


is  the  nth.  derivative  of  f{x),  that  is,  the  result  of  differen- 
tiating fix)  n times.  The  following  notation  is  also  used, 

-£f(x)  =f'(x),  j*-J{x)=f\x),  -,  £-nf(x)=f^(x). 

The  operation  of  finding  the  successive  derivatives  is 
called  successive  differentiation. 


EXERCISE  9 

1.  Given  f(x)  = 3 x4  — 4 x2  6 a*  — 1, 

then  fix)  = 12  — S x + 6 ; 

f"(x)  = 36  jr2  — 8,  etc. 
el.  calc.  — 4 


50 


DIFFERENTIATION 


2.  Given 

3.  Given 

4.  Given 

5.  Given 

6.  Given 


f(x)=eax\  prove  f(n\x)  = aneaz. 
f(x)  = loge(i  - x)  ; prove  fM(x)  = 


y — x3  loge  x\  prove 


d*y  _ 6 
dx*  x 


(—  i)n|7z  — i 
0 *)* 


, . r , d3y  2 cos  x 

y — l°ge  sin  x;  find  ^ = 


y = e 2x ( x 2 — 3 + 3)  ; find 


dzy 

dx3 


7.  Given  -b  + ys  — z*  or  b2x2  + a2y2  = aW,  to  find 


d2y 


b 2 

From  Ex.  2,  Exercise  8, 

^ = _ 

dx 


bbc_ 

ay 


d-y 
dx 2 


dx2 


a*y^x)~  Vx^ay 


d2y 

dx2 


a2b2y  — b2a-x‘^- 
a*y2  ’ 


dy 

then  substituting  for  and  reducing, 

d2y  _ b2(a'y+b2x2)  _ b* 

a2ys 

d2y  _ 4 fi2 


dx2 


8.  From 


y2  = 4 fix,  find  ^4  = = - S 


9.  From  y2  — 2 xy  = a2,  prove 


dx 

d2y  _ a2 
dx2  — (y  — x)3 


xy 


CHAPTER  III 


APPLICATIONS 


21.  Tangent  and  Normal.  For  all  applications  of  the 
Calculus  to  Geometry  the  fact  established  in  § 14  is  of 
fundamental  importance,  viz. 

Theorem.  The  value  of  the  derivative  of  y with  respect 
to  x found  from  the  equation  of  a curve  in  rectangular  coor- 
dinates gives  the  slope  of  the  tangent  at  any  point  on  that 
curve , or 

— = slope  of  tangent. 
dx 

If  we  wish  the  slope  at  any  particular  point  (x',y'),  we 
have  to  substitute  x'  and  y'  respectively  for  x and  y in  the 

general  expression  for  — • Let  (~-\  be  the  value  of 
, dx  \dx) 

— after  this  substitution,  then  we  have  from  Analytic 
dx 

Geometry, 

Equation  of  the  tangent  at  (x1 , y')  is 


(1 7) 


-v'=(d£)[x-x')- 


y-y 


Since  the  normal  is  perpendicular  to  the  tangent,  and 
dy\ 


from  ( 1 1 ),  [-£)  = 7^7,  we  find 


Eqtiation  of  the  normal  at  (pc’ , y')  is 
(18) 

51 


y-y’  = - (^j  (x  - x'). 


52 


APPLICATIONS 


EXERCISE  10 


1.  Find  equations  of  tangent  and  normal  to  the  parabola  /2  = 4 x+ 1 
at  the  point  whose  ordinate  is  3. 

Substituting  3 for/,  we  find  x = 2,  hence  (pc' , /')  is  (2,  3).  Differen- 
..  ..  dy  2 . /^/Y  2 

t,atm&»=7  ■■[■£)  =-3- 

Ans.  tangent,  7.  x — 3/  + 5 = 0;  normal,  3 jr  + 2/  — 12  =0. 


2.  Find  equation  of  tangent  to  the  ellipse  b^x1  + a2/2  = aW  at 

(x',  /').  ^u'br-)-  a^y'/  = a2^2. 

3.  Show  (Fig.  12)  that  the  subtangent  MYTX  =—  y'(  — \ , and  the 

, , \ ' \dy  1 

subnormal  MXNX  = /'  \-d-  \ • 


4.  Prove  that  the  subnormal  in  the  parabola  /2  = 4 px  has  the  con- 
stant length  2 p. 


22.  Sign  of  the  Derivative.  An  important  question  is 
the  following : 

Is  the  function  increasing  or  decreasing  as  the  variable 
passes  through  a given  value  a ? 

The  phrase  “ passing  through  a ” is  understood  to  mean 
that  the  series  of  values  assumed  by  the  variable  is  an 


increasing  sequence  including  a,  i.e.  on  the  graph  of  the 
function  we  proceed  from  left  to  right.  In  Fig.  12,  as  we 


APPLICATIONS 


53 


pass  through  Px  the  ordinates  are  decreasing,  while  at  P2 
the  ordinates  are  increasing,  and  since  the  ordinates  repre- 


of  the  tangent,  we  have  the  result : 

The  function  f{x)  is  increasing  or  decreasing  as  x passes 
through  a according  as  f(a ) is  greater  or  less  than  zero. 

At  P3  and  Pi  (Fig.  12)  the  tangent  is  parallel  to  XX1,  and  therefore 
f'(x)  vanishes  at  these  points.  For  such  values  of  x,  therefore,  the 
rule  just  given  does  not  enable  us  to  answer  the  question  proposed. 

If,  now,  for  any  value  of  x,  say  x=a,  the  second  deriva- 
tive l~t,,  or  f"(x),  is  positive,  then  as  ,r  passes  through  a, 
dx- 

the  first  derivative  f\x),  or  tan  7,  must  be  an  increasing 
function  of  x,  i.e.  7 must  be  increasing  as  x passes  through 
a ; and  therefore  as  we  pass  along  the  curve  from  left  to 
right,  the  tangent  is  rotating  counter-clockwise,  and  the 
curve  is  accordingly  concave  upward  (as  at  (a),  Fig.  13). 


On  the  contrary,  if  f"(d)  < o,  the  reasoning  shows  the 
tangent  to  be  rotating  clockivise  as  we  pass  along  the  graph 
through  x — a,  and  hence  the  curve  is  concave  downward 
((f),  Fig.  13). 


Fig.  13 


54 


APPLICATIONS 


The  result  is : 

A curve  is  concave  upward  or  downward  as  x passes 
through  a according  as  the  value  of  the  second  derivative 

r , , 

-jx 2 for  x = a is  greater  or  less  than  zero, 
dfy 

As  before,  if  = o,  the  rule  just  given  does  not  enable 

us  to  decide.  If  = o for  x = a and  changes  sign  as  *- 

passes  through  a,  then  at  x = a we  have  a point  of  inflec- 
tion (Aj  and  Pb,  Fig.  12). 


exercise  11 

1.  Show  that  the  following  functions  are  either  always  increasing  or 
always  decreasing,  and  draw  the  graphs  in  each  case : 

(a)  tan*;  (b)  (c)  log*;  (d) 

2.  Show  that  y = sin*  has  a point  of  inflection  at  each  intersection 
with  XX1. 

3.  Determine  the  points  of  inflection  of  y = (*  — a)s  + b. 

Atis.  ( a , b). 

23.  Maxima  and  Minima.  A function  fix)  is  said  to  be 
a maximum  for  x = a when  f(a)  is  the  greatest  value  of 
fix')  as  *-  passes  through  a. 

A function  fix)  is  said  to  be  a minimum  for  x = a when 
fia)  is  the  least  value  of  fix)  as  **  passes  through  a. 

In  other  words,  a maximum  value  is  greater  than  any 
other  in  the  immediate  vicinity,  and  similarly  for  a mini- 
mum value.  It  is  not  to  be  inferred  that  a maximum  value 
is  the  greatest  of  all  values  of  the  function ; on  the  con- 
trary, a function  may  have  several  maxima. 


APPLICATIONS 


55 


Graphically,  at  a maximum  we  have  a highest  point 
(. Pl  and  P3,  Fig.  14),  at  a minimum  a lowest  point 
\p 2 and  Pfi 


Fig.  14 


Since,  by  definition,  if  f(a)  is  a maximum,  f(x)  must  be 
an  increasing  function  for  x < a and  a decreasing  function 
for  x>  a,  we  have  (§  22): 

Theorem.  If  f(a ) is  a maximum  value  of  fix),  then 
the  first  derivative  fix)  must  change  sign  from  positive  to 
negative  as  x passes  through  a. 

By  similar  reasoning  for  a minimum,  we  find  a change 
in  sign  from  negative  to  positive  must  occur  in  fix). 

In  either  case,  therefore, /'(hr)  must  change  sign.  If  we 
now  assume  that  fix)  is  continuotis  for  x — a,  we  see  that 
f(a)  = o ; that  is,  the  tangent  at  a highest  or  lowest  point 
must  be  horizontal  and  P2  in  Fig.  14).  If,  on  the  con- 
trary, fix)  is  not  continuous  for  x = a,  then  the  change  in 
sign  occurs  by  passage  through  co  ; i.e.  the  tangent  becomes 
parallel  to  YY' , as  at  P3  and  P4.  This  case  is,  however, 
of  minor  importance,  and  is  omitted  from  further  con- 
sideration. 

Furthermore,  if  f\a)<o,  the  curve  at  ;tr  = a is  concave 
downward,  and  we  have  a highest  point  {P-f,  while 
f'(a)  > o indicates  a lowest  point  (P2). 


56 


APPLICATIONS 


We  have  therefore  the  following 

Rule  for  determination  of  Maximum  and  Minimum 
values  of  a function  fix). 

Find  the  first  derivative  fix),  and  get  the  roots  of  the 
equation  f'( x)  = 0. 

First  Test.  If  f'(x)  changes  sign  as  x passes  through 
any  root  a of  the  equation  f'(x)  = 0,  then  f(a)  is  a maxi- 
mum or  minimum  value  according  as  the  change  is  from  + 
to  — , or  from  — to  + . 

Second  Test.  Find  the  second  derivative  f'(x) ; then, 
if  a is  any  root  of  f'(x)  = 0.  f(a)  is  a maximum  if  f"  (a)  < 0. 
and  a minimum  if  f"  (a)  > 0.  If,  however,  f'(a)  = 0,  we 
must  use  the  first  test. 


examples 


1 . Examine  the  function  x3  — 3 x3  — 9 x + 5 for  maxima  and  minima. 

Placing  f{x)=x3  — 3 x2—  gx+  5, 

then  f'(x)=3x2—6x—g, 

and  the  roots  of  3 x3—  6x—  9 = 0 are  x=  3 and  — 1. 

Now  f"(x)  = 6x—6,  and  f"  (3)  = 12,  /"(—  1)  = — 12, 
hence  by  the  Rule,  Second  Test, 

f(2>)  — — 22  is  a minimum  value, 
and  y(  — 1)  = 10  is  a maximum  value  of  the  function. 

The  student  should  draw  the  graph. 


2.  Examine  the  function  ^ for  maxima  and  minima. 
(x+  i)3 


Here 

Differentiating  and  reducing,  we  find 


r,  v (X  — i)2 

fix)  - !—■ 

J y ’ (x+  I)3 


f (x)  (^  — 1 ) 6r  ~ 0- 

J K 1 ix+  iy 


APPLICATIONS 


57 


The  roots  of  /'  (x)  = o are  therefore  x=  I,  x=  5.  We  now  apply  the 
First  Test,  since  it  is  unwise  to  form  the  second  derivative. 

Taking  account  of  the  signs  only,  we  have 


When*x<  1,  f'(x)  = — £ — ^ — — 
When  x>  1,  /'(x)  = — ^ — - 


= + 


Hence  f{x)  is  a 
minimum  when  x = 1 . 


When  x < 5,  /'(x)  = - ^ = + 1 , . . 

y + I Hence  /'  (x)  is  a 

,Trl  _ n , \ ( + )(  + ) f maximum  when  x=  c. 

When  x>5,  f\x)~-  ’ = - j ^ 

/(i)  = o is  a minimum,  and  /(5)  = ^ a maximum  value  of  the 
function. 


EXERCISE  12 


1.  Examine  the  following  functions  for  maxima  and  minima : 

(*) 

x2—  3 x+  5. 

Minimum  value  1 1 . Ans. 

(*) 

X 

Max.  value  min.  value  — \.  Ans. 

I +x2 

(0 

6 x+  3 x'1—  4 x3. 

Max.  value  5,  min.  value  — J.  Ans. 

(d) 

x3—  3 x2+  6x. 

No  max.  or  min.  values.  Ans. 

0) 

<2X2+  2 bx  + c. 

If  a > 0,  min.  value  — — — , if  a < 0, 

a 


then  — — — is  a maximum.  Ans. 
a 

(/")  iov/8  x — x2.  Max.  value  40.  Ans. 

This  function  is  a maximum  or  minimum  according  as  8 x — x-  is 
a maximum  or  minimum,  hence  f a constant  factor  or  a radical  sign 
may  be  dropped  in  investigations  of  this  sort. 

* We  consider  values  of  x differing  only  very  slightly  from  the  number  on 
the  right  of  the  inequality  sign. 

t If  u is  any  polynomial  in  x containing  710  multiple  factors,  we  may  show  that 
fu  is  a maximum  or  minimum  only  when  u is  a maximum  or  minimum.  For  if 


58 


APPLICATIONS 


2.  Divide  the  number  a into  two  such  parts  that  their  product  shall 
be  a maximum. 

Hint.  If  ur  is  one  part,  then  a — x is  the  other,  and  the  function  to 
be  examined  is  x(a  — x)  or  ax  — x~.  Equal  parts.  Ans. 

3.  Divide  the  number  a into  two  such  parts  that  the  product  of  the 
in th  power  of  one  and  the  «th  power  of  the  other  shall  be  a maximum. 

In  the  ratio  m : n.  Ans. 

24.  The  subject  of  Maxima  and  Minima  is  one  of  the 
most  important  in  the  applications  of  the  Calculus  to  Ge- 
ometry, Mechanics,  etc.  It  is  often  necessary  to  derive 
the  expression  for  the  function  to  be  investigated,  and  in 
testing  this,  attention  should  be  paid  to  the  remark  in 
Example  i (/)  of  the  preceding  exercise. 


EXERCISE  13 


1.  A box  with  a square  base  and  open  top  is  to  be  constructed  to 
contain  108  cubic  inches.  What  must  be  its  dimensions  to  require  the 
least  material  ?*  Base  6 inches  square,  height  3 inches.  Ans. 


2.  The  strength  of  a rectangular  beam  varies  as  the  product  of  the 
breadth  b and  the  square  of  the  depth  d.  What  are  the  dimensions  of 
the  strongest  beam  that  can  be  cut  from  a log  whose  cross  section  is  a 
circle  a inches  in  diameter  ? f Breadth  is  % a inches.  Ans. 


H — so  that  fix') 


f(x)  = Va,  fix)  = — - — — , and  f"(x)  = — - — 1 


vanishes  only  if  — = o,  and  then  f"  ( x ) has  the  same  sign  as 

dx  dx 2 


108 


* HINT.  Let  JV  be  the  side  of  the  base,  y the  height,  then  x^y  ~ 108,  i.e.  y - 

x- 

and  since  the  material  is  x2  + 4 xy,  we  find  by  substituting  for  y the  function 
*2  + 432. 


t Hint.  The  strength  therefore  equals  W2  multiplied  by  some  constant,  which 
may  be  dropped  by  the  remark  of  § 24.  But  d2  = a2  — ^2;  hence  the  function  is 
1 5 (a2  — b2) , b being  the  variable. 


APPLICATIONS 


59 


3.  Find  the  dimensions  of  the  stiffest  beam  that  can  be  cut  from 
the  same  log  as  in  2,  given  that  the  stiffness  varies  as  the  product  of 
the  breadth  and  the  cube  of  the  depth.  Breadth  \ a inches.  Ans. 


4.  The  equation  of  the  path  of  a projectile  (see  Fig.  16)  is 

,-2 


y = x tan  a. 


gx< 


2 VS  COS"  « 


where  a is  the  angle  of  elevation  and  v0  the  initial  velocity.  Find  the 

greatest  height 


v0z  sin  2 u 


Ans. 


5.  Find  the  dimensions  of  the  rectangle  of  greatest  area  that  can  be 
inscribed  in  the  ellipse  b2x2-\-a2y'2  — a2b2.  Ans.  Sides  are  aV 2 and  bV2. 

6.  Find  the  altitude  of  the  right  cylinder  of  greatest  volume  inscribed 

in  a sphere  of  radius  r.  Altitude  = IT.  Ans. 

v3 


6o 


APPLICATIONS 


7.  Assuming  that  the  brightness  of  the  illumination  of  a surface 
varies  directly  as  the  sine  of  the  angle  under  which  the  light  strikes  the 
surface  and  inversely  as  the 
square  of  the  distance  from  the 
source  of  light,  find  the  height 
of  a light  placed  directly  over 
the  center  of  a circle  of  radius 
a when  the  illumination  of  the 
circumference  is  greatest. 

From  Fig.  17,  the  bright- 
ness at  P is  given  by 

k sin  0 kx 


X 

\d 



— - 

Fig.  17 


d2 


d 3 


V ( a 2 + x2) 
Hence  the  brightness  is  a maximum  when 


•2\3  ( (a2  + 


,(a2  + *2: 

x'2 

(d2  + x'2)3 


is  a maximum. 


.*•= . Ans. 

V2 


25.  Expansion  of  Functions.  By  actual  division 


— I “f"  X -f-  X -f-  • • • -f-  Xn  -j” 


09) 

V ' I — X 

where  n is  some  positive  integer.  In  this  simple  way  we 

may  find  for  the  function  1 an  equivalent  polynomial 

all  of  whose  coefficients  save  that  of  xn+l  are  constants.  By 
transposition  (19)  becomes 


(20) 


I — a: 


— ( I + x + x1  + A"3  + •••  + xn)  = 


I — x 


„7!+l 


Now  let  x be  some  number  numerically  less  than  1,  say 
x=.5,  and  suppose  we  wish  the  value  of  correct 

within  one  one-hundredth,  i.e.  correct  to  two  decimal  places. 
Let  us  then  determine  for  what  values  of  n the  term 
1 


1 —x 

equality 


xn+l  when  x=.$  is  less  than  .01,  i.e.  solve  the  in- 


1 

1 - -5 


• 5"+1<.oi.  We  find  n>6. 


APPLICATIONS 


6l 


Furthermore,  if  ;r  is  numerically  less  than  .5,  

1 — x 

and  xn+1  are  less  than  for  .r  = .5,  so  that  taking  n — 7 
(J.e.  >6),  ■ — ~x8<  ,oi  for  every  value  of  a- not  numeri- 
cally greater  than  .5.  And  we  now  see  from  (20)  that 
the  function  ^ 1 may  be  replaced  by  the  polynomial 

I + x + x2  + x3  + x*  + x5  + x6  + x~  for  all  values  of 
numerically  equal  to  or  less  than  .5  if  results  correct  only 
to  hundredths’  place  are  desired. 

Precisely  the  same  reasoning  holds  for  any  value  of  x 
numerically  less  than  unity,  since  for  any  such  value  jtrn+1 
can  be  made  as  small  as  we  please  by  taking  n sufficiently 
great.  But  this  reasoning  does  not  hold  for  any  value 
equal  to  or  exceeding  1 numerically.  We  may  then  state 
this  theorem : 

For  any  value  of  x numerically  less  than  unity , the  func- 
tion — - — may  be  represented  with  any  desired  degree  of 

accuracy  by  a sufficiently  great  number  of  terms  of  the 
polynomial, 

I + x + x2  + x3  + •••. 

The  Differential  Calculus  enables  us  to  obtain  a similar 
theorem  for  many  other  functions,  as  will  now  be  explained. 
In  all  practical  computations  results  correct  to  a certain 
number  of  decimal  places  are  sought,  and  since  the  process 
in  question  replaces  a function  perhaps  difficult  to  calcu- 
late by  a polynomial  with  constant  coefficients,  it  is  there- 
fore of  great  practical  importance  in  simplifying  such 
computations. 


62 


APPLICATIONS 


26.  Theorem  of  the  Mean.  If  f(x)  and  f(x)  are  con- 
tinuous as  x varies  from  a to  b,  then  there  is  at  least  one 
value  of  x,  say  xv  between  a and  b,  such  that 


(21) 


f(b)-f(a) 
b — a 


=f(* 1). 


In  Fig.  11,  f(b)—f(a)=CB,  b — a = AC, 
A6)-/(a) 


b — a 


= slope  of  AB,  and  at  each  of  the  points 


Px  and  P 2 the  tangent  is  parallel  to  AB,  and  hence  (21)  is 
true  if  xx  is  the  abscissa  of  Px  or  P2.* 

Multiplying  (21)  out  gives 


(22)  /(b)  =f(a)  + (b  - a)f'(x1), 

where  it  must  be  remembered  a >xx  > b. 

A more  general  theorem  than  (21)  is  enunciated  as 
follows : 

If  f(x)  and  the  (;/  + 1)  successive  derivatives  f'(x), 
f"(x),  f(v+1)(x)  are  continuous  when  x varies  from  a 


* This  proof  of  the  theorem  of  the  mean  is  not  mathematically  rigorous,  but 
merely  illuminates  the  significance  of  (21).  The  student  should  draw  other  figures, 
and  especially  such  that  the  necessary  conditions  of  the  continuity  of  f[x)  and 
f'{x)  fail. 


APPLICATIONS 


63 


to  b,  then  there  is  at  least  one  value  of  ;r,  say  xv  between 
a and  b such  that 

(23)  /W=/W  + (-^/0)  + ^|^/"W 

+ (-^j /"(?)+  -+  (-^r/<'>(«) 


(*  - a) 
\»+  1 


n+ 1 


The  proof  of  (23)  is  beyond  the  scope  of  this  book.* 
The  student  should,  however,  carefully  note  the  law  by 
which  the  expression  on  the  right  is  constructed. 

Putting  for  b in  (23)  the  variable  -r,  we  get  Taylor s 
Theorem , 

(24)  f(x)  = f (a)  + ----- /*(«)  + (-|2-)2/^)  + ••• 

(x — aY+x  . n/  . 

+ — / 1 (-^1),  where  a<x\  < x. 

\n  + 1 1 L 


Finally,  setting  a = o in  (24),  we  find  Maclaurin  s 
Theorem, 

(25)  /M=/(°)+f/'(o)+jf/"(o)  + - +L/“<°) 


+ 


X* 


+1 


71  + I 


f(n+l\xx)  where,  o <x1<  x. 


If  in  (23)  we  put  b — a + x,  we  obtain  another  form  of  Taylor’s 
theorem, 

f(a  + x)  =f(a)  + + ...  etc. 

This  formula  (25)  gives  f{x)  in  the  form  of  a polynomial 
in  .r  with  constant  coefficients  save  that  of  xn+l,  which, 
since  xl  lies  between  o and  x,  is  a function  of  x ; that  is, 


* An  excellent  discussion  is  given  in  Gibson’s  An  Elementary  Treatise  on  the 
Calculus,  London,  1901,  p.  390. 


C4 


APPLICATIONS 


we  have  the  generalization  of  the  example  of  § 25  as 
follows : 

A function  f (x)  for  certain  values  of  the  variable  * may 
be  represented  with  any  desired  degree  of  accuracy  by  the 
polynomial, 

/(o)+f/(o)  +|>(o)  +jy/"(o)+  ...  +jf /(">(o). 

By  “ expansion  of  a function  ” is  meant  the  forming  of 
this  polynomial.  Of  course  71  is  indefinite,  and  must  be 
taken  great  enough  to  give  the  desired  degree  of  accuracy. 
It  is  of  greatest  theoretical  importance  to  determine  for 
what  values  of  x the  polynomial  represents  the  function 
when  n is  taken  indefinitely  great.  This  consists  in  exam- 
ining for  what  values  of  x 


Limit 


V \n  + 1 


for  this  term  is  the  difference  between  the  function  and  the 
polynomial. 

EXERCISE  14 

1.  Expand  sin  x. 

Since  f(x)  = sin  x,  and  for  x = o,  /(o)  = sin  o = o ; 

then  f(x)  = cos  x,  and  for  x = o,  /'( o)  = 1 ; 

f"  (r)  = — sin  x,  f (o)  = o ; 

f"(x)  = — COS  X,  f"(o)  = — I i 

fn(x)  = sin  x,  ffo)  = o ; 

etc.  etc. 

x3  x5  x~  x9 

Hence  sin  x=  x — , 1- , . 1- . etc. 

[3  LL  IZ.  12. 

2.  Show  that  the  expansion  of  cos  a-  is 

X2  X4  X*  Xs 

COSX=  I b . ,-w  + 1-5-  - etc- 

h Li  II  11. 


* Namely,  for  all  values  of  x such  that  the  11  remainder” 


2-n+l 

T+T 


/["+« (^i) 


is 


less  than  the  limit  of  error.  This  question  is  often  difficult  to  settle. 


APPLICATIONS 


65 


3.  Expand  ex. 

Since  f(x)  = e x,  and  all  its  derivatives  are  likewise  ez,  while  e° 
we  obtain  x *2  *s  ^4  ^5 

e T = 1 H 1- 1 h . h , hi h • • • • 

■ ll  ll  [4  [5_ 

Putting  x = 1,  we  find 


1 1 1 1 

e — I H h, h, h, h 

1 ll  ll  1 4 


h 


the  expression  given  in  § 10. 

The  expansions  of  sin  x,  cos  x,  and  ex  are  remarkable  in  that  they 
hold  for  every  value  of  x,  positive  and  negative. 


4.  Prove  the  following  expansions 

l-2  -v-3 


(v  y2  -y-O  y-4  yO  \ 

T-T  + y-T  + y--) 

(£)  (1  + x)m  = i + mx  + 


m(m  ~ 1 „2 

II.  13 


**  + W(W  - l)(w-2)^  + 


(£)  is  the  binomial  formula.  These  expansions  hold  only  for  values 
of  x numerically  less  than  1 . 


Taylor's  Theorem  (24)  differs  from  (25)  in  that  we  are 
to  consider  values  of  the  variable  x near  some  given  num- 
ber a , since  (24)  is  a polynomial  in  (x  — a)  in  the  same 
sense  that  (25)  is  a polynomial  in  x.  It  is  evident  that  no 
greater  difficulty  arises  in  the  application  of  (24)  to  a given 
function  than  has  been  already  pointed  out. 


27.  Differentials.  From  (23)  we  are  able  to  find  an  ex- 
pansion for  the  increment  of  a function  in  powers  of  the 
increment  of  the  variable  as  follows  : 

Write  b = x + Ax,  a — x,  ,\  b — a = Ax,  and  (23)  be- 
comes, after  transposing  fix), 

(26)  fix  + Ax)  -fix)  = Ax f{x)  + ^p-/'  \x)  + • • •, 
or  (27)  A fix)  = fix)  Ax  +f"ix) 


Now,  if  we  suppose  Ax  to  diminish  toward  zero,  the  first 
term  fix)  Ax  of  the  right-hand  member  will  ultimately 
el.  calc.  — 5 


66 


APPLICATIONS 


greatly  exceed  the  sum  of  the  remaining  terms,  since  these 
contain  higher  powers  of  Aw.  For  this  reason  fix)  Aw  is 
called  the  principal  part  of  the  increment  of  fix).  Also, 
when  we  wish  to  emphasize  the  fact  that  the  variable  Aw 
is  to  approach  zero  as  a limit,  we  write  dx,  called  differen- 
tial x,  instead  of  Aw,  and  the  principal  part  of  the  incre- 
ment f(x)dx  we  call  the  differential  of  the  function;  that  is, 

(28)  d fix)  = fix)  dx. 

The  following  definitions  are  fundamental: 

A differential  (or  infinitesimal)  is  a variable  whose  limit 
is  zero. 

The  differential  of  the  independent  variable  is  an  incre- 
ment of  that  variable  whose  limit  is  zero. 

The  differential  of  the  dependent  variable  is  the  princi- 
pal part  of  the  increment  of  that  variable,  and  equals  the 
product  of  the  derivative  and  the  differential  of  the  inde- 
pendent variable  (28). 

From  (28),  we  see  that  if  y is  a function  of  w,  then 

(29)  dy  = ^dx. 


EXERCISE  15 

1.  Prove  by  (28)  and  (29)  the  following  differentials : 
(a)  d (3  w2)  = £>xdx. 

dx 

if  d loge  x = — • 


dx 


(e)  d Vi  — x = — ■ 

2 V 1 — x 

(f)  d sin  2 x = 2 cos  2 xdx. 

(f)  dez  — eTdx.  sec2 

(d)  dxm  = mxm~^dx.  d tan  ( ..'j  — pi 

(h)  If  y = wlogew,  then  dy  =(1  + logew)^w. 

2.  If  / = uv , then 

dy  = ( u — - + v \ dx  = u — dx  + v — dx,  or  dy  = udv  + v du. 
\ dx  dx / dx  dx 


dx. 


APPLICATIONS 


67 


3.  Show  that 


!0- 


v du  — u dv 
v2 


4.  State  the  rules  I-V  for  differentiation  in  terms  of  differentials 
instead  of  derivatives. 


28.  We  may  write  (27)  after  replacing  A,r  by  dx, 

(30)  A f{x)  =f(x)dx  + dx2  + pj-'  dx+~ ^j. 

Now,  since  by  (28)  f{x)dx  is  the  differential  of  the 
function,  (30)  shows  that  A fix)  and  df{x)  differ  by  a term 
containing  the  factor  dx2.  Such  a quantity  is  called  a 
differential  of  the  second  order ; in  general,  any  quantity 
containing  as  a factor  the  product  of  tzuo  differentials  is 
thus  designated. 

The  increment  of  a function  differs  from  its  differential 
by  a differential  of  the  second  order. 


EXAMPLES 

1.  Differential  of  a product  uv. 


Let  u — AB,  v = AC,  then  uv  = area  ABCD.  If  du  = BB’, 
dv  = CC,  then 

A {uv)  = area  AB'  CD’  — area  ABCD 

= area  CDC' E + area  BB'DE'  + area  DE' ED' 

= u dv  + v du  + du  • dv. 

Now  du  ■ dv  is  a differential  of  the  second  order,  .-.  principal  part 
of  A (uv)  is  udv  + vdu;  i.e.  d{iiv')  = udv  + v du.  (Cf.  Ex.  2,  § 27.) 


68 


APPLICATIONS 


2. 


Differential  of  an  area. 


Consider  the  area  aAPM  bounded  by  any  curve,  the  axis  XX'  and 
the  ordinates  aA,  MP,  and  call  this  area  u.  Then  if  MN  = dx,  A u 
= axtaaAQN— area  aAPAI=  zrta MPQN.  \u  = y dx  + area PSQ. 

But  area  PSQ  < dx- dy.  .-.  PSQ  = k dxdy,  where  k is  some  number  < i. 
Hence  area  PSQ  is  a differential  of  the  second  order,  and  du=ydx. 

The  differential  of  the  area  bounded  by  any  curve , the  axis  XX',  and 
two  ordinates  is  the  product  of  the  ordinate  of  the  curve  and  the  differ- 
ential of  the  abscissa. 


3.  Differential  of  the  volume  of  a solid  of  revolution. 

Let  the  solid  be  generated 
by  revolving  a curve  APQ 
around  XX',  and  denote  the 
volume  APA'P  by  v.  If 
dx=  MX,  then  \v  = volume 
AQA'Q'  — volume  APA'P', 
or  Sv  = volume  of  the  cylin- 
der PSP'S'  + volume  gener- 
ated by  the  curvilinear  A PSQ. 

Now  the  volume  of  the  cylin- 
der PSP'S'  = iry'-dx,  since  y 
— PM  = radius  of  base  and 
dx  = altitude.  The  volume 
generated  by  the  curvilinear 
A PSQ  < volume  generated 
by  the  rectangle  PRSQ,  and 
this  last  volume  = ttNQ~  • MN  — ttMP ' • MN  — tt(2  y dy  -f  dyfdx. 


Fig.  2i 


APPLICATIONS  69 

We  see  therefore  that  Av  = 7ry2  dx  + a differential  of  the  second 
order,  i.e.  dv  = iry-dx. 

The  differential  of  the  volume  of  a solid  of  revolution  generated  by 
revolving  any  curve  around  the  axis  XX'  equals  n tunes  the  product  of 
the  square  of  the  ordinate  and  the  differential  of  the  abscissa. 


4.  Show  that  the  differential  of  the  area  71  bounded  by  a curve  AP 
and  two  radii  vectores  OA  and  OP  is  given  by  du  — \ r2dO,  where 
(r,  6)  are  the  polar  coordinates  of  P. 


CHAPTER  IV 


INTEGRATION 


29.  Indefinite  Integral.  Integration  consists  in  finding 
a function  of  which  a given  differential  expression,  such  as 

x dx,  sin  x dx,  — , etc.,  is  the  differential.  The  function 

n 

thus  found  is  called  the  integral  of  the  given  differential 
expression,  and  the  operation  is  indicated  by  prefixing  the 

integral  sign  Thus,  since 


J'x  dx  = 


1 j-2  . 

O --V  i 


d (|^  x2)  = x dx, 

J* dx  = x,  ^ sin  x dx  = — cos  x,  etc. 

In  general, 

§ fix)  dx 

means  to  find  a function  F( x)  such  that 
dF(x)  = f{x)dx, 


i.e. 


Ax)=f*F(x)- 


Constant  of  Integration.  Since  d( ^ x2  + C ) also  equals 
a'  dx,  no  matter  what  the  constant  C is,  we  have 


^ x dx  = jX2  + C, 


where  C is  any  constant  whatever,  called  the  constant  of 

:foi 
7° 


integration.  We  see,  therefore,  that  a given  differential 


INTEGRATION 


71 


expression  may  have  infinitely  many  integrals,  found  by 
giving  to  the  constant  of  integration  different  values. 
Thus 

J' f{x)  dx  = F{x)  + C, 

and  since  C is  unknown  and  indefinite , F(x ) + C is  called 
the  indefinite  integral  of  f{x)dx. 

Of  course,  the  same  differential  expression  has  an  in- 
definite number  of  distinct  integrals,  but  what  has  just 
been  said  shows  that  the  difference  of  any  two  of  these 
must  be  a constant. 


30.  Rules  for  Integration.  From  Rule  V in  differentia- 
tion, if  v is  any  function  of  ;v,  and  k a constant,  then 

dL(KV\  = k—,  i.e.  dinv)  = k dv. 
dx  dx 

Integrating,  we  have,  since  if  two  differential  expressions 
are  equal  so  are  their  integrals  equal, 


j" k dv  = j* d{KV), 

or,  since 

J*d(/cv)  = KV, 

kv  =J^/cdv. 

But 

*f dv  = KV. 

(30 

.’.  J' k dv  — kJ ' dv. 

XXIII.  A constant  factor  may  be  written  either  before  or  after 
the  integral  sign. 

The  chief  application  of  XXIII  is  to  be  found  in  cases  like  the 
following : 


72 


INTEGRATION 


To  work  out  J* xdx . If  we  multiply  xdx  by  2,  we  have  an  exact 
differential,  since 

d (x2)  = 2 xdx, 


but  by  XXIII, 


J*2  xdx  — x 2 ; 
2 xdx  — 2 xdx, 


-i 

From  (31)  we  may  also  write 


xdx=  — 
2 


(32) 


ff(x)dx=~  j%Kf(x)dx 


Integral  of  a Sum  of  Differential  Expressions.  If  u and  v 
are  functions  of  x,  then 

d(u  + v)  = — (11  4-  v)  dx  = du  + dv. 
dx 

J*(du  + dv)  =j*d(u  + v)  = u + v = j' du  +J'dv. 


This  result  gives  Rule 

XXIV.  The  integral  of  any  algebraic  sum  of  differential  ex- 
pressions equals  the  same  algebraic  sum  of  the  integrals  of  these 
expressions  taken  separately. 

That  is,  eg., 

j (x  + 3)  d x=  (xdx  + 3 dx)  = xdx  + j3dx=%x2+3x  + C. 

31.  From  any  result  in  differentiation  may  always  be 
derived  an  integration  formula,  and  we  now  proceed  to 
obtain  some  of  the  simpler  ones,  making  use  of  § 18. 


INTEGRATION 


73 


Since  by  VIII, 

d(vm+1)  = (in  + i )vm  dv, 

then,  integrating, 


m + i)vmdv  = ( in  + i 


(XXIII) 


(33) 

From  IX, 


vm+l 

m + i 


dlogev  = 


dv 

5 

V 


(34) 


In  the  same  way  we  might  go  through  with  each  formula 
in  § 1 8.  It  will  suffice  for  our  purpose  to  tabulate  a few 
of  the  results : 


XXV. 

f vm  civ  = v>n  1 + C ( m d - 
J m + 1 

XXVI. 

f^  = loge  v + C. 
J V 

XXVII. 

^avclv  = °v  +C. 
loge  a 

XXVIII. 

y sin  v dv  = — cos  v + C. 

XXIX. 

^ cos  v dv  = sin  v + C. 

XXX. 

f dv  — arc  sin  v + C. 
' d a-  — v 2 a 

XXXI. 

f ,^-  = 1 arctan^  + C. 
J a%  + v%  a a 

74 


INTEGRATION 


1.  Find 


EXAMPLES 

dx 


S' 


V i — x 

This  is  the  same  thing  as  J*  (i  — x)~7dx,  which  resembles  XXV. 
For  put 

i — x — v,  then  — dx  = dv,  or  dx  = — dv. 

J*(i  — x)  2 dx  = ^v  7 — dv  = — ^v  -dv. 


by  XXV 

and  by  substituting  again, 


T v 2 dv  = — + C, 

J i 


f — —x  ■ — — 2 Vi  — x + C. 
J V i — x 


2.  Work  out 


Si 


3 ax  dx 

¥72 


Taking  out  constant  factor  3 a (XXIII),  this  becomes 

x dx 


C x dx 

3>a  \ tt 

J c-  — b-x 


and  this  resembles  XXVI. 


dv 


For  put  c 2 — b2x2  = v,.-.  — 2 b2xdx  = dv,  or  xdx  — 

2 b- 

_ dv 

r xdx  r id1  3 a C dv  3 a , ^ 

J c2  — b2x 2 j v 2 b- J v 2 b'2 

C ?>axdx  = _ 3_^  log  (cl  _ b2x2)  + C. 

J c2  - b2x2  2 b2  J 


3.  Find 


S 


dx 

9 + 4 X2' 


This  resembles  XXXI,  if  a = 3,  2 x = v. 

Then  2 dx  = dv,  and  since  the  given  integral  by  (32)  is  the  same  as 

1 C z dx  or  I r dv_  t 

2 J 32  + (2  x)2  2 J a2  + v2 

we  find  by  XXXI,  f — — — = - arc  tan  — + C. 

J 9 + \x2  6 3 


INTEGRATION 


75 


By  studying  the  above  examples  the  student  will  see 
that  integration  depends  upon  comparison  of  the  given 
integral  with  certain  standard  forms.  To  be  able  to  tell 
quickly  what  form  the  given  integral  resembles  is  abso- 
lutely essential. 

Tables  of  standard  forms*  have  been  constructed  con- 
taining all  integrals  occurring  in  ordinary  work. 


EXERCISE  16 

1.  Prove  the  following  integrations  : 

{a)  J {ax  + bx2)  dx=\ax2  + \ bx 8 + C.  (Use  XXIV.) 


sin  xdx 


= logc  ( i - cos  x)  + C. 


(*>  /“ 

(c)  J y/a2  - ar2  xdx  _ _ i(a2  _ xiy  + c.  (Use  XXV,  v = a2  - x2). 
W iO»3X+C. 


(^)  y £-x  dx  — — e-*  + C. 


<£)  S 


(/>  y 


dx 


xdx 
Va2+x 
dx 


'•  Va2  + x2f  C. 


V i — 4 . 


1 arc  sin  (2  x)  + C.  {h)  f— h-  = -logTi -ar)  + C. 

2 J i — ar 


(0  y ^ xdx  — §xi  + C ; + C. 

J J X°  2 X* 

C sin  x 

(j)  i tan  xdx  = - logc  cos  a-  + C.  (Put  tan  ar  = and  use 

J ° x cos  ar 

XXVI.) 

(k)  fan*  xdx  =\x-  f sin2ar  + C.  (Put  sin2  ar  = |(i  - cos  2 ar).) 
2.  Special  Devices  in  Integration. 

(a)  By  partial  fractions , when  we  have  to  integrate  a rational  frac- 
tion times  dx,  and  this  fraction  can  be  replaced  by  partial  fractions. 


E.g. , B.  O.  Peirce’s  A Short  Table  of  Integrals,  Ginn  & Co.,  1899. 


76 


For  example, 


INTEGRATION 

dx 


JV 


i A B 

a2  — x2  ~ a — x a + x 


2 a 


Putting 

and  clearing  of  fractions, 

i = x{A  — B)  + a(A  + B). 

.-.  A — B = o,  a(A  + B)  = i,  or  A = B = 

=i>G^D+c 

(b)  By  change  of  variable. 

Find  J" V a2  — x2 dx.  Substitute  x=azosQ\ 


dx  = — a sin  6 dd,  y/a2  — x2  — fa2  — a2  cos2  6 = a sin  0, 
and  J*  y/a2  — x2  dx  = — a2j  sin2  8dQ  = — ^-0  + — sin  20  + C 


by  Ex.  i (£).  Now 


0 = arc  cos  sin  2 0 = 2 sin  8 cos  6 


= 2y[i 


•••  i’V^7 


x'2  dx  = arc  cos  - + - xV a2  — x2. 

2 a 2 


3.  Prove 


J; 


<aJr 


= los:c  — — — + C. 


jr2  + 3 ar  ^&e\x  + 3 
The  following  two  examples  illustrate  the  manner  of  determination  of 
the  constant  of  integration  by  means  of  so-called  initial  conditions. 

4.  Find  the  amount  of  a sum  of  money  increasing  continuously  at 
compound  interest  of  r per  cent. 

We  found,  page  43,  that,  in  derivatives,  P being  the  sum  sought, 

— = — P. 
dt  100 

Multiplying  by  dt  and  dividing  by  P,  we  have 

— = -6-  dt ; 

P 100 


too 


t -f-  C. 


integrating, 


(0  log£/>  = 


INTEGRATION 


77 


Let  now  a equal  the  initial  sum  of  money ; that  is,  the  sum  started 
with,  so  that  P = a when  t = o ; substituting  these  in  this  equation, 

we  have  logea=C,  so  that  (i)  becomes  log,, P = — t + loge a,  or, 


transposing, 


I Of 


tP  - logea  = — t,  or  loge(-)  = — 
ioo  \a  J ioo 


P = aem  . Ans. 


5.  Find  the  relation  between  s (space)  and  t (time)  for  uniformly 
accelerated  rectilinear  motion. 

dv  d'l) 

Since  the  acceleration  — is  constant,  say  f,  we  have  — = f. 

dt  dt 

Multiplying  by  dt , dv  = f dt , and  integrating,  v —ft  + C. 

To  determine  C,  let  the  initial  velocity  be  v0,  i.e.  v = v0  when  t — o, 
or  v0  — o+C.  v =ft  + v0. 

Since  v — ft  + v<v  and  multiplying  by  dt,  ds=ft dt  + vudt. 

Integrating,  s — \ft2  + vat  + C,  and  if  s = j-0  when  t = o,  we  have 
finally  s = \fP  + vj  + j0.  Ans. 


32.  Definite  Integral.  We  have  already  seen  that  the 
indefinite  integral  contains  an  arbitrary  constant,  the  con- 
stant of  integration,  and  has  for  that  reason  an  indefinite 
value.  By  making  suitable  assumptions,  now  to  be  ex- 
plained, we  are  able  to  dispose  of  this  inconvenience. 

In  § 2 8,  Example  2,  it  was  shown  that  the  differential  of 

the  area  u between  a 
curve  MABC,  the  axis 
XX' , and  any  two  ordi- 
nates was  given  by 
dn  — y dx. 

/.  u = dx  + C. 

Here,  of  course,  y is 
some  function  of  x determined  from  the  equation  of  the 

curve,  and  .\  § y dx  =■  some  function  of  x,  say  F(x). 
u = F(x)  + C. 


78 


INTEGRATION 


Let  us  now  agree  to  reckon  the  area  from  the  axis  YY\ 
so  that  when  x = a,  u = area  OaAM,  etc. 

Under  this  assumption,  when  x = o,  u — o,  and 

o=  F(o)  + C,  or  C = — F(o), 

and  we  have 

u — F{x)  — F(o). 

Now  area  OaAM  = F{a)  — F(o). 

Area  ObBM  — F{b ) — F{o).  Subtracting,  we  have 

Area  abAB  = F{b )—  F(a), 
or, 

The  difference  of  values  of  the  \ y dx  for  x — b and  x = a 

gives  the  area  bounded  by  the  curve  zvhose  ordinate  is  y,  the 
axis  XX' , and  the  ordinates  at  a and  b. 

This  difference  is  represented  by  the  symbol 

(35)  jaydx’ 

read,  “ integral  from  a to  b of  y dx" ; the  operation  is  called 
integration  between  limits , a being  the  lower,  b the  upper 
limit. 

We  see  therefore  that  (35)  or,  what  is  the  same  thing, 

(36)  jaf(x)dx 

always  has  a defiiiite  value,  and  is  accordingly  a definite 
integral.  For  if 

(37)  Jf{x)dx  = F(x)  + C,  then 

(38)  jj(x)dx  = F(b)  + C-  (F(a)  + C)  = F(b)  - F(a), 
and  the  constant  of  integration  has  disappeared. 


INTEGRATION 


79 


33.  Areas  of  Plane  Curves.  From  § 32,  we  have  the 

theorem:  Given  any  plane  curve  y = f{x),  the  definite 

integral  f f(x)dx  gives  the  area  bounded  by  that  curve, 

the  axis  XX'  and  the  ordinates  at  a and  b. 

To  find  the  area  bounded  by  two  given  curves,  we  get 
the  area  between  each  and  XX'  and  then  subtract. 

Volumes  of  solids  of  revolution. 

Precisely  as  in  § 32  and  remembering  the  result  of 
Example  3,  § 28  we  prove  that : 

Given  any  plane  curve  y — f(x),  the  definite  integral 
J"  Try2  dx  gives  the  volume  generated  by  revolving  around 

XX'  the  portion  of  the  curve  between  the  ordinates  at  a 
and  b. 

The  two  theorems  just  given  find  numerous  applications 
in  Geometry. 

EXERCISE  17 

1.  Find  the  area  of  the  curve  y = x2  — 9 lying  below  XX' . 

Here  y dx  = ^ (x2  — 9)  dx,  and  since  for  y = 0,  x=  ± 3,  the  limits 
rs 

are  +3  and  —3,  i.e.  area=J  fix2—  f)dx.  36.  Ans. 

2.  Find  the  area  of  the  circle  x2  + y2  = a2. 

Since  y = da2  —x2,  ^ ydx  = ^ da2—  x2 dx  which  has  been  worked 
out  in  Exercise  16,  Example  2 ( 'b ).  For  the  semicircle  the  limits 
are  + a and  — a. 

3.  Show  that  the  area  of  the  ellipse  b2x2y  a2y2  = a2b2  is  to  the  area 
of  the  circle  whose  diameter  is  the  major  axis  2 a as  b : a. 

4.  Find  the  area  of  one  arch  of  sine  curve  y = sin  x.  2.  Ans. 

5.  Find  area  between  the  equilateral  hyperbola  xy  = 1.  the  axis 
XX',  and  the  ordinates  at  x=  a,  x = b. 


l0&(;)- 


Ans. 


8o 


INTEGRATION 


6.  Find  the  volume  of  the  sphere. 

Since  we  have  to  revolve  the  circle  x2+y2  = a2,  or  y2  = a2  — x2 
around  XX',  then  J* -rry2dx  = i rj*  (a2  — x2)  dx.  The  limits  are  + a and 
— a.  f-7 ra3.  Ans. 

7.  Find  the  volume  generated  by  revolving  around  XX'  the  pa- 

rabola /2  = 4^rand  cut  off  by  a plane  perpendicular  to  XX'  at  the 
distance  of  4 to  the  right  of  the  origin.  32  -rr.  Ans. 


34.  Definite  Integral  as  the  Limit  of  a Sum  of  Differential 
Expressions.  In  the  Differential  Calculus  the  student  was 
asked  to  bear  in  mind  that  everything  was  built  up  from  a 
fundamental  limit,  the  limit  of  a quotient  whose  denominator 
approached  zero.  We  are  now  to  see  that  the  definite  inte- 
gral is  the  limit  of  a sum  of  differential  expressions. 

If  j' f{x)dx  = F(x)  + C, 

then  —F(x)  = f(x)  and  f f{x)dx  =F(b)  — F(a) 
dx  Ja 

gives  the  area  bounded  by  the  curve  y =f{x)  (Fig.  24), 
the  axis  XX' , and  the  ordinates  at  ^'=  a,  x = b. 


Now  divide  the  segment  ab  into  any  number  of  equal 
parts,  say  6,  albl  = bf2  = •••  =bf,  and  call  the  length  of 
each  division  A^r.  Erect  the  ordinates  at  these  points,  and 


INTEGRATION 


8l 


apply  the  theorem  of  the  mean  (§  26)  to  each  division. 
In  the  present  case  Fix')  takes  the  place  of  fix)  in  (21), 
and  fix)  replaces  f\x );  for  the  first  interval  abv  a— a, 
b = bv  and;r1(  lying  between  a and  b,  is  marked  in  the  figure. 
Draw  the  ordinate  of  xv  Then  (21)  gives 

Fjb^-Fja) 

b,-a 

or,  since  bx  — a = Ax, 

(39)  Fib1)-Fia)=fix1)  Ax. 

In  the  same  way  (21)  applied  to  each  of  the  remaining 
five  segments  gives  the  equations 


(40) 


Fibz)  - Fib\)  =fi*f)Ax, 
F(Z,3)-Fib2)=fix3)Ax, 

Fibi)  - Fibs)  =fix 4)A-G 

Fib5)~Fibi)=AX^X> 
Fib)  -Fib5)=fix6)Ax. 


Adding  the  six  equations  (39)  and  (40),  we  find 
(41 ) Fib)  - F{a)  =fixx)Ax  +f(x2)Ax  + /( xs)Ax 

+fixi)&x  +fix5)&x  F fix^jA.r. 


But  fix-^Ax  = area  of  the  rectangle  aPPxbv 
fix2) Ax  = area  of  the  rectangle  b-J)xP2b2, 
etc., 


so  that  the  sum  on  the  right  equals  the  area 

aPPxpxP2p2Pzp3P,p,PhphQb ; t.e. 

(42)  Fib)  —F{a)  = area  between  the  broken  line 

FFiPi-PnQ  and  XX', 

EL.  CALC.  — 6 


82 


INTEGRATION 


and  this  is  true  independently  of  the  number  of  parts  into 
which  ab  is  divided.  Hence  for  any  number  n of  equal 
parts 

(43)  Fib)  —F(a)  = f{x^)\x  +f(x2)kx  4 +f{xn) Ax, 


(44) 


and  A.r  = 


b — a 


Equations  (43)  and  (44)  hold  when  n increases  without 
limit,  and  then  A;r  becomes  dx  (§  27),  i.e.  a variable  whose 
limit  is  zero. 


F(b)—  F(a)=  L™*  (f(xl)dx+f(x2)dx+ ...  +f(xn)dx), 


or,  by  (38), 


(45)  fa  f(x)dx  = (f(x1)dx+f(x2)dx+ ...  4 -f{xn)dx). 


And  now  we  see 


very  clearly  why  I f(x)dx  gives  the 

d 


area  under  th£  curve,  for  as  n increases,  the  broken  line 
PPxPxP^p^  •••  pbQ  approaches  the  curve  itself,  and  the  sum 
fix-P)dx  + -..  f(xn)dx  always  represents  the  area  under 
this  broken  line. 

Integrating  between  limits  is  accordingly  spoken  of  as 


“summing  up”;  the  integration  sign  j is  historically  a 

distorted  S,  the  first  letter  of  swn.  But  let  the  student 
not  forget  that  the  definite  integral  is  not  a sum,  but  the 
limit  of  a sum , the  number  of  terms  increasing  without 
limit , and  each  term  itself  diminishing  toward  zero. 

The  problem  of  finding  the  area  is  then  to  be  thought 
of  thus : Divide  the  interval  on  xx'  into  an}'  number  of 
equal  parts,  and  at  a point  within  each  division  erect  an 
ordinate  to  the  curve ; construct  the  rectangles  on  the 
divisions  as  bases,  with  the  corresponding  ordinate  as 


INTEGRATION 


83 


altitude.  Then  finding  the  area  consists  in  summing  up 
these  rectangles  and  taking  the  limit  of  this  sum  as  the 
number  of  divisions  increases  without  limit. 


As  an  example  of  the  great  number  of  problems  in  Physics  and  other 
branches  of  Mathematics  which  involve  in  their  solution  definite  inte- 
grals, consider  the  following : 

To  determine  the  amount  of  attraction  exerted  by  a thin,  straight, 
homogeneous  rod  of  uniform  thickness  and  of  length  l upon  a material 
point  P of  mass  in , situated  in  the  line  of  direction  of  the  rod. 


dx  - 


Fig.  25 

Imagine  the  rod  (see  Fig.  25)  divided  up  into  equal  infinitesimal 
portions  (elements)  of  length  dx.  If  M = mass  of  rod,  then 

-jdx  — mass  of  any  element. 

The  law  of  attraction  being  Newton’s  Law,  i.e.  attraction  = product 
of  masses  h-  square  of  distance,  then 

M 


idx 


attraction  of  element  dx  on  P = ■ 


(x+a)2 

and  the  total  attraction  is  the  sum  of  these  from  x = o to  x — l. 

M 


Force 


= f 

-Jo 


mdx 

1 l Mm  n dx 


(x+a)2  l Jo  (x+a)2 

or  integrating,  Force  = ( 1 f d \ = — . Answer. 

5 l \ 1+ a a)  a(a  + l) 


CHAPTER  V 


PARTIAL  DERIVATIVES 

35.  Functions  of  More  than  One  Variable.  In  the  pre- 
ceding chapters  we  have  been  concerned  with  functions  of 
one  variable ; i.e.  the  variable  function  depended  for  its 
value  upon  the  value  of  a single  variable.  Such  functions 
do  not  by  any  means  suffice  for  the  applications  of  the 
Calculus.  In  fact,  the  student  is  already  familiar  with 
many  examples  of  a variable  whose  value  depends  upon 
those  assigned  to  two  or  more  distinct  variables.  Thus 
the  area  of  a rectangle  is  a function  of  two  variables,  viz. 
the  two  sides ; the  volume  of  a gas  depends  upon  both  the 
pressure  and  the  temperature ; the  volume  of  a parallelo- 
piped  depends  upon  the  three  edges,  etc. 

Notation.  If  the  value  of  a variable  u depends  upon 
two  variables,  x and  y,  and  can  be  computed  when  values 
are  assumed  for  x and  y,  then  we  write  precisely  as  in  § 3, 

(46)  u =f(x,  y). 

Similarly  for  a function  of  three  variables, 
u = cf)(x,  y,  2),  etc. 

36.  Partial  Differentiation.  As  in  § 12  the  important 
question  arising  here  is  how  to  determine  the  manner  of 
variation  of  the  function  when  the  variables  change  in 
value.  But  we  have  greater  latitude  here  than  in  § 12. 
For  in  (46)  we  can  ask  ourselves, 

84 


PARTIAL  DERIVATIVES  85 


first , how  does  n vary  when  ;r  alone  varies  and  y remains 
constant  ? or 

second , how  does  it  change  when  remains  constant  and 
1 y varies  ? or 

third,  in  what  manner  does  u vary  when  both  and  y 
change  independently  of  each  other  ? 

Thus  let  u-  xy,  x and  y being  respectively  the  base  and  altitude 
of  a rectangle ; if  y remains  constant  (say  y = b),  11  gives  the  area  of 
all  rectangles  of  a certain  altitude  b ; and  if  x = a constant,  say  a , then 
u represents  the  area  of  all  rectangles  with  common  base  a.  But  if 
x and  y both  vary  independently,  then  we  are  to  consider  all  possible 
rectangles. 

Now  the  first  and  second  cases  do  not  differ  in  the  least 
from  § 12,  for  we  really  have  in  the  first,  u a function  of  x 
alone,  and  in  the  second,  u a function  of  y alone.  We  can 
therefore  form, 

first,  the  increment  quotient  (§  13)  when  x alone  varies, 
and  this  is 


(47) 


An  _ f{x  + Ax,  y)  — f(x,  y) _ 

Ax  Ax 


second , the  increment  quotient  when  y alone  varies,  which 
is 

r.st  Au  - f('r>  y + Ar)  -f(x,  y) 

(4§)  Ay  Ay 

For  example,  in  the  area  of  rectangle  already  used,  u — xy, 

A u (x  + Ax)v  — xy  ' A u . 

-t—  = t — = y,  and  ---  reduces  to  x. 

Ar  A.r  J A y 

Finally,  we  can,  as  in  § 14,  find  the  limits  of  the  func- 
tions in  the  right-hand  members  of  (47)  and  (48),  in  (47) 
when  Ax  approaches  zero,  in  (48)  when  Ay  approaches 


86 


PARTIAL  DERIVATIVES 


zero.  The  results  are  called  the  partial  derivatives  of  u 
or  f[x,  y)  with  respect  to  .v  and  y respectively,  and  this 
step  of  passing  to  the  limit  we  indicate  on  the  left  by 
replacing  the  A’s  by  round  d’s,  so  that 


(49) 


du  T . . 
— — Limit 
ox 


ffx  + Ax,y)-f(x,y)\ 

Ax  Ai=o’ 


(50) 


du  T . 

— = Limit 
dy 


7(l  y + A v)  ~/(l  j>A 

Ay  Jav= o’ 


The  partial  derivatives  — , — are  then  to  be  calculated 

dx  dy 

by  the  rules  of  Chapter  II,  the  independent  variables  being 
respectively  ;r  and  y. 


EXERCISE  18 


1.  Find  the  partial  derivatives  of : 


0 * = iog,(r). 

(2)  rc  = arc  tan  - 

(3)  u = xv. 


A us. 

du  _ 

1 

dx 

X 

Ans. 

du  _ 

dx 

X2 

A ns. 

du  _ 
dx 

yx y- 

du  _ 
' dy~ 

y_ . 

2+y2' 

i • §11 

dy 


y 

du 

dy 


x 2 +y2 


Partial  Differe7itials : 

By  § 27,  (29),  the  differential  of  u,  when  x alone  varies  is 

— dx,  and  when  y alone  varies  equals  — dy ; these  are 
dx  " dy 

called  the  partial  differentials  of  u. 


C r\ 

— dx  = partial  differential  of  u,  when  x alone  varies ; 
I dx 

r) 

— dy  = partial  differential  of  u,  when  y alone  varies. 
dy 


PARTIAL  DERIVATIVES 


s; 


37.  Total  Differentiation.  We  have  yet  to  discuss  the  third  case 
of  § 36,  viz.  required  the  change  in  u when  .rand  / vary  independently. 
If  Ar,  A/,  and  Az<  are  the  increments  of  these  variables,  then  from  (46) 
we  have 

(52)  A u = /(r  + Ar,  / + Ay)  -f(x,  y). 

By  adding  and  subtracting  f(x,  y + Ay)  in  the  right-hand  member, 
(52)  becomes 


(53)  Aa=/(r+Ar,  y + Ay)  -/(r,  y + Ay)  +f(x,  y + Ay) -f(x,  y). 
Consider  now  the  last  two  terms, 

f(x,y  + Ay)  — f(x,  y). 

This  is  the  increment  of  u or  f (x,  y)  when  / alone  varies.  Hence, 

by  (27),  § 27, 

(54)  f (+7  + A y)  ~ = Ay  + terms  in  higher  powers  of  Ay. 

In  the  same  way  the  first  two  terms  of  (53)  give  us,  if  we  set 


u'  =/(r,  y + Ay), 

(55)  /0+  Ar,/  + Ay)  -f(x,y  + Ay) 


But  also 


= dff  Ax+  terms  involving  Ar2,  etc. 
dx 


it!  = f{x,y  + Ay)  = /(r,/)  + |"A/  + terms  in  A/2,  etc., 


= — + terms  in  Ay, 
dx  dx 


by  (26),  § 27.  Differentiating  with  respect  to  r,  we  find 

(56) 

since  H —f(X,  /). 

Consequently,  from  (56),  (55),  and  (54)'  (S3)  becomes 

/ c 7\  Au  = — Ar  + — Ay  + terms  of  higher  degree  in  Ar,  Ay. 

> dx  dy 

Now  letting  Ar  and  Ay  approach  zero,  i.e.  become  the  infinitesimals 
dx  and  dy , then,  as  in  § 27,  calling  the  principal  part  of  Au  the  total 
differential  of  u,  we  have 

(58)  du  = Yxclx  + dydy' 


88 


PARTIAL  DERIVATIVES 


From  (51)  and  (58),  then,  we  have  the  theorem: 

The  total  differential  of  a function  of  several  variables 
equals  the  sum  of  the  partial  differentials. 

Example.  In  § 28,  Example  1,  was  demonstrated  the  result 
d ( xy ) = xdy  + y dx, 

which  agrees  with  (58). 

EXERCISE  19 

Find  the  total  differentials  of  the  following : 

(a)  u = log,(£).  A ns.  du  = 

/ y\  xdy  — y dx 

{b)  u = arc  tan  1 - 1 . Ans.  du  = ^ + a — 

( c ) u = xy.  Ans.  du  = xi'~1(y  dx  + xloge  xdy). 


38.  Total  Derivative.  We  may  in  (57)  assume  that 
and  y are  not  independent,  but  are  functions  one  of  the 
other,  say  y a function  of  x.  Then  u becomes  also  a func- 
tion of  ;r  alone,  and  we  may  therefore  form  the  total 

derivative  — - 
dx 

Dividing  (57)  by  Ax  and  taking  the  limit  for  A.r  = o,  and 
.•.  Ay  = o,  we  have  the  result 


(59) 


du  _ du  _ f du\  dy 
dx  dx  \dyj  dx 


a very  important  formula. 


Suppose  in  the  illustration  of  the  rectangle,  § 36,  we  wish  the  deriva- 
tive with  respect  to  the  base  x of  the  area  u of  all  rectangles  whose 
altitude  / is  double  the  base.  Then 


du  du  dy 

u = xy,  y = 2 x,  ~=y,  3-  = *,  fL  = 2> 
dx  dy  dx 


du 

dx 


= y + 2x=4x. 


and  (59)  gives 


PARTIAL  DERIVATIVES 


89 


Or,  we  may  substitute  for  / before  differentiation; 

i.e.  u = x • 2x  = 2x2,  .-.  — = \x,  as  before. 

dx 

Equation  (59)  is  especially  important  as  affording  a proof  of  the 
method  given  in  § 19.  For  in  the  example  of  that  article,  set 

u = ar2  _ 3 xy  + 2y-2  _ 3 . 


i.e.  (60) 


u = o,  and 


du 

dx 


o,  or 


du  + (3u\^  = 0. 

dx  \dy / dx 


du  ' 

dy  _ dx  _ 2x—t,  y _2x—  3/ 

dx  du~  —3  a- +4/  2>x  - 4/’ 

dy 


the  same  answer  as  before.  This  formula  (60)  is  very  useful. 


For  further  study  of  the  Calculus  the  student  is  referred  to  : 

G.  A.  Gibson,  An  Elementary  Treatise  on  the  Calculus.  London, 
1901. 

Young  and  Linebarger,  The  Eleme7its  of  the  Differential  and 
Integral  Calculus.  New  York,  1900. 

McMahon  and  Snyder,  Elements  of  the  Differe7itial  Calculus. 
New  York,  1898. 

Murray.  A71  Eleme7itary  Course  m the  I/itegral  Calculus-.  New 
York,  1898. 


CHAPTER  VI 


ADDITIONAL  EXAMPLES  IN  CONNECTION  WITH 
THE  GIVEN  EXERCISES 


EXERCISE  1.  PAGE  10 

6.  Given  f(pc)  — x3  — 6 x2  + 5 ; 

then  by  § 3,  /( 2)  = (2)*  - 6(2)'2  + 5 

= 8 — 24  + 5 = — n. 

Similarly,  /(-  3)  = (~  3)3  - 6(  - 3)2  + 5 

= - 27  - 54  + 3 = - 76. 

Also  f(x  — t)  = (x  — i)3  — 6(x  — i)'2  4-  5, 

which  reduces  to  = x3  — 9 x2  + 15  .r  — 2. 

7.  Given  = 3 x3  — 2 x2  + 6 x,  prove 

A-(o)=o;  /■(!)=  7;  F(-  2)=  - 44; 

F(i  -y)  = 7 - 11  y + 7 y2  - 37s- 


8.  Given 


% _i_  i 

<f>(x)  = , prove  the  following: 


x — i 


<K 2)  = 3 ; <t>(~  i)—  ° ; <^>(i+^)  = 


2 + X 


9.  In  Example  7,  prove  </ > 


y± 
y - 


EXERCISE  2.  PAGE  22 

X \ / 

6.  Draw  the  graphs  of  51 ; 1 Vi—x2;  csc.r;  cot2t 

7.  Prove  the  following : 

Limit  resell  =00;  Limit  P All  =00:  Limit  + 1 1 

L Lx-iJ, 


90 


ADDITIONAL  EXAMPLES 


91 


EXERCISE  3.  PAGE  24 


It  is  often  convenient  to  place  the  variable  y equal  to  the  given  func- 
tion of  x.  Then  when  x changes  to  x + h,  y is  replaced  by  y + A/, 
that  is,  A y is  the  increment  of  the  function.  The  following  examples 
illustrate  this,  and  in  them  the  increment  of  x is  denoted  by  A.r  instead 
of  h,  as  heretofore. 


9.  Given  y = x"1  — 3^+6;  find  A y. 

Substituting  x + tS.x  and  y + Ay  for  x and  y respectively,  we  have 
y + Ay  = (x  y A.r)'2  — 3(x+  Ax ) + 6 

= x2  y 2 x-  Ax  + (Ax)2  — 3 x — 3 Ax  + 6. 
Subtracting  from  this  the  original  equation,  we  find 
Ay  = 2 x • A.r  + (A.r)2  — 3 Ax, 


or 

Ay  = (2  x — 3 + Aar)  A.r.  Ans. 

10.  Given 

y = x3;  find  Ay. 

Ay  = (3  x2  + 3 • A.r  + (A.r)2)  Ax.  Ans. 

11.  Given 

y = 1 ^ 'x  ; find  A_y. 

x 


* x2  — i + x • A x \ . 

Ay  = ! Ax.  Ans. 

x2  + x ■ Ax 

12.  Given  y = -^r ; find  Ay. 

y/x 

Ay  = — — Ax.  Ans. 

(x  + Ax)  Vx  + xy/x  + Ax 


EXERCISE  4.  PAGES  32-33 


As  above,  if  y is  set  equal  to  the  given  function  of  x to  be  differ- 
entiated, the  General  Rule  of  Differentiation  (page  29)  may  be  stated 
as  follows  in  Four  Steps : 


1st  step. 
2d  step. 

3d  step. 
4th  step. 


Substitute  sc  + Ax  and  y + Ay  for  x and  y,  respectively. 
Subtract  the  original  equation  from  the  new  equation,  and 
reduce. 

Divide  both  sides  of  this  equation  by  Ax. 

Find  the  limit  of  this  result  when  Ax  approaches  the  limit 

zero,  replacing  ||  by 


92 


ADDITIONAL  EXAMPLES 


9.  Differentiate  j = 3 j2  + 2J  + 5. 


Carrying  out  the  Four  Steps  : 

i°.  Substituting  x + Ax  and  / + A/  for  x and/, 

/ + A/  = 3(x  + Ax)2  + 2 (x  + Ax)  + 5, 
or  / + Ay  = 3 x2  -f  6 x - Ax  + 3 (Ax)2  + 2 x + 2 Ax  + 5. 

20.  Subtracting  the  original  equation  and  factoring,  we  have 
A/  = (6  x + 3 Ax  + 2)  Ax. 

30.  Dividing  by  Ax, 

Ay 

= 6 x + 3 Ax  + 2. 

Ax 

. i\y  dy 

4°.  Letting  Ax  approach  the  limit  zero  and  replacing  by 

4^  = 6x  + 2.  Ans. 

dx 

X 

10.  Differentiate  / = 3- 

y 1 + x2 


Working  out  the  Four  Steps  gives : 
x+  Ax 


reducing, 

3°- 


y + A/  = 

A/  = 
A/=- 


1 + (x  Ax)  ■ 

x + Ax  x 

1 + (x  + Ax)2  1 + x'2 
1 — x2  — x ■ Ax 


(1  + x2)(i  + (x  + Ax)2) 
Ay  1 — x2  — x • Ax 

Ax-  (1  + x2)(i  +(x+  Ax)2) 
dy  _ 1 — 3 


Ax. 


Ans. 


* ' dx  ( 1 + x2)2 

11.  Differentiate  y = }x3-x2-  3X+  5. 

^ = x2  — 2 x — 3.  Ans. 
dx  J 


12.  Find  the  slope  of  the  tangent  to  the  curve/  = £x®  - x2  - 3 x+  5 
at  the  points  where  x = o,  x=3,  x=  — 1,  x = 1. 

From  Example  1 1 and  page  27,  the  general  expression  for  the  slope 


at  any  point  is  given  by  the  value  of  i.e. 


slope  = x2  — 2 x — 3. 


ADDITIONAL  EXAMPLES 


93 


Hence,  substituting, 

slope  at  x — o is  — 3 ; 
slope  at  x = 3 is  9 — 6 — 3=0; 
slope  at  x—  — 1 is  1 + 2 — 3=  0; 
slope  at  x=  1 is  1 — 2 — 3 = — 4. 

13.  Find  the  general  expression  for  the  slope  of  the  tangent  to  each 
of  the  following  curves  : 

(a)  y = xi  — 8 x2  + 16.  4 x3  — 16  x. 

2 x 8 — 2 x2 


V)  y = 

(0  y = 


4 + x2 
8 a3 


(4  + x2)2' 

— 1 6 a3x 


Atis. 
A ns. 


A ns. 


x2  + 4 a2  (x2  + 4 a'2)  2' 

14.  Determine  the  coordinates  of  all  points  on  each  of  the  curves  of 
Example  13  at  which  the  tangent  is  parallel  to  the  axis  of  x. 

Hint.  Since  at  these  points  the  slope  of  the  tangent  is  zero,  place 
the  general  expression  for  the  slope  equal  to  zero,  and  solve  for  ar. 

{a)  (o,  16),  (±2,0);  (b)  (2,  i),  (-2,  (c)  (o,  2 a).  Ans. 


EXERCISE  8.  PAGE  48 

dy  x — y 

8.  Given  x2  — 2 xy  = a2 ; prove  ~ = — — • 

9.  Given  ( x — a)2  + (y  — b)2  = r2\  prove  ^ = — ^ ^ 


10.  Given  x2  + y2  = a2  ; prove 


dy_ 

dx 

dr 


11.  Given  r 2 sin  2 6 = a2 ; prove  — - = — r cot  2 0. 

do 

12.  Given  the  curve/3  = ; show  that  the  axis  of  / is  the  tangent  at 

the  origin. 

EXERCISE  9.  PAGE  49 

If  the  variable  / is  set  equal  to  the  function  of  x to  be  differentiated, 

then  the  first  differentiation  gives  the  value  of  -]-■  The  second  de- 

& dx 

rivative,  obtained  by  differentiation  from  this,  is  represented,  as  in 

d2y 

§ 20,  by  (read,  “the  second  derivative  of  / with  respect  to  x"). 


94 


ADDITIONAL  EXAMPLES 


Then  further  successive  differentiation  will  give  the  third,  fourth,  fifth 

d^y  dAy  dhy 

derivative,  and  so  on,  which  are  symbolically  ^4,  etc. 

d4y 

10.  Given  y — e~z  cos  x ; prove  — — 4/. 

d‘2y  2 X 

11.  Given  / = arc  tan  .r ; prove  = — 


dx2  ( 1 + x'1)2 

1 d2y  1 — x ^ 

12.  Given  / = loge  (1  + x2)2  ; prove  — 


diy . 


dx2  (1  + x:)2 


13.  Given  y = xcos  x;  prove  —4  = x cos  •*■  + 4 sin  ar. 

. d2y 

14.  Given  / = sin2  2 ;r ; prove  = 8 cos  4 x. 


d2y  cl 2 

15.  Given  x2  + 2 xy  — a1 ; prove  ^ 


d^y 


16.  Given  y = tan2  x + 8 log  cos  x + 3 x2 ; prove  -j~  — 6 tan4  x. 

r.  *3  d4y  [4 

17.  Given / = — ; ^ = (7^' 


d3y 


4 a 3 


18.  Given  / = (x2  + a2)  arc  tan  - ; prove  ^ ^2- 

d^y  — 24.  x 

19.  Given  y2  + y = x2  ; prove  -dr,  = — — 

7 7 r dx 3 (i  + 2/)5 

„ d2y  a(m2  — 1) 

20.  Given  y2  — 2 m xy  + x2  — a = o ; prove  -yy.  = -A ~ 

y ' ’ r dx1  (/  — mx) 3 

;r  d2y  2x(x2-3) 

21.  Given  / = s ; prove  -yy.  = 

7 1 + x2  ' F dx2  (1  + a.-2)3 


EXERCISE  10.  PAGE  52 

5.  Find  equations  of  tangents  and  normals  to  the  curve  y2  — 
2 x2  — xa  at  x = 1 . Alls.  Tangents  are  / = 4 x + ^ ; / = — 1-  x — 

Normals  are  / = — 2 x + 3;  y = 2 x — 3. 

6.  Prove  that  subtangent  and  subnormal  to  the  ellipse  b2x2  + a2y 2 

oP> x'2  b^x1 

= a2b2  at  (xr,  /')  are  — -p- — and  — respectively.  (Use  formulae 
of  Example  3.) 


ADDITIONAL  EXAMPLES 


95 


7.  Find  equations  of  tangents  and  normals  to  2 x'2  + 3/2=35 

at  / — 3.  Ans.  Tangents  are  4.x±  9 / = 35. 

Normals  are  9 a-  T 4/  = 6. 

8.  Find  equations  of  tangent  and  normal  to  / = x3  at  (x',  /'). 

Ans.  Tangent  is  3 x'2x  — y — 2 y'  = o. 

Normal  is  x + 3 x"2y  — x'(3  x '4  + 1)  = o. 

X 

9.  Find  equations  of  tangents  and  normals  to  y = — , at  = 1 . 

Ans.  Tangents  are  y = ± l- 
Normals  are  x = ± 1. 

10.  Prove  that  the  equation  of  the  tangent  to  x 3 — 3 axy  + y3  = 0 
at  {x',  /')  may  be  written  x"2x  — ax’y  — ay' x +y'2y  = o. 


EXERCISE  11.  PAGE  54 


4.  Prove  that  the  curve/ = 
and  x = ± aVy 


x 

a 2 + :r2 


has  points  of  inflection  at  x=  o 


5.  Examine  the  curve  a2y  = i x3  — ax2  +2  a3  for  points  of  inflection. 

Ans.  ( a , | a) . 


6.  Examine  y = x + 36  x2  — 2 x3  — x*  for  points  of  inflection. 

Ans.  At  x = 2,  x = — 3. 


EXERCISE  12.  PAGE  57 

Referring  to  Fig.  14,  page  55,  we  see  that  the  conditions  for  maximum 
and  minimum  values  of  a function  may  be  stated  as  follows,  the  graph 
of  the  function  being  supposed  drawn  : 

Maximum : tangent  horizontal  and  graph  below  the  tangent  (e.g.  at  P^) ; 

Minimum : tangent  horizontal  and  graph  above  the  tangent  (^.at  Pg). 

If  we  set  y equal  to  the  given  function  of  x,  then  the  preceding  may 
be  stated  thus : 

Maximum:  and  ^^<0: 

ax  dx1 

Minimum  : / - = o and  > o. 
dx  dx 2 


96 


ADDITIONAL  EXAMPLES 


This  is  identical  with  the  Second  Test,  page  56,  and  the  process  is 
summarized  in  the  following  rule  : 

1st  step.  Set  y equal  to  the  given  function  of  x and  work  out 

„ dy  , d-y 
expressions  for  ^ and 

2d  step.  Set  the  expression  found  for  equal  to  zero,  and  solve 

A (lOO 

for  x. 

3d  step.  Substitute  each  value  of  x thus  found  in  the  expression 

If 

for  ^5.  If  this  result  is  negative,  the  corresponding 

value  of  y is  a maximum ; if  positive,  the  correspond- 
ing value  of  y is  a minimum. 

To  illustrate,  solve  the  problem : To  find  the  maximum  and  minimum 
values 

2 X 

Following  the  rule,  we  have 
-f-  4 

i°.  Assume/  = then,  differentiating  and  reducing, 

dy  _ x2  — 4 . 
dx  ~ 2 x'1  ’ . 


d2y  _ 4 
dx2  x8 


Setting  the  expression 


dx 


or 

30.  If  x — 2,  then 


x2  - 4 = 0, 

x = ± 2. 

d ~y  _ 4 _ r . 
dx 2 2 3 2 ' 


if  ur=  —2,  then 
Therefore,  since 

and 


d2y  _ 4 _ 1 

dx2  — 23  2 

x = 2 gives  y — 2, 
x = — 2 gives  y — — 2, 


then  2 is  a minimum  value,  and  — 2 a maximum  value  of  the  given 
function. 


ADDITIONAL  EXAMPLES 


97 


4.  Find  maximum  and  minimum  ordinates  in  each  of  the  following 
curves,  and  draw  the  loci : 


(a)  _y  = 2 x3  — 2 x2  — 12  x + 4. 

Ans.  Max.  value  n,  min.  value  - 16. 


(£)  y = 2 X3  — 21  X2  + 36 

x — 20. 

Ans.  Max.  value  — 3,  min.  value  — 128. 

6 

+ 

* 

N 

1 

H 

II 

Vsi 

Ans.  Max.  value  10,  min.  values  ± 9. 

(d)  y = (x  — l)3(x  — 2)“. 

Ans.  Max.  value  min.  value  0. 

H' 

° H 
II 

Ans.  Max.  value  -• 
e 

( f ) y = sin  2 x — x. 

Ans.  Max.  values  when  x = (n  + £)tt; 

Min.  values  when  x = (n  — -J)7r, 

n being  any  integer. 

, N sin  x 

■’'  = ,+  tan  * 

Ans.  Max.  value  \ V2. 

(/i)  y = sin  x(  I + cos  x). 

Ans.  Max.  value  |V3,  min.  value  — 5V3. 

(0  y = x*  - 4. 

Ans.  Min.  value  — 4. 

(j)  y = xs-  8. 

Ans.  No  max.  nor  min.  values. 

CO* 

1 

II 

Ans.  Max.  value  4. 

EXERCISE  13.  PAGE  58 


8.  Find  the  altitude  of  the  cone  of  maximum  volume  which  can  be 

inscribed  in  a sphere  of  radius  r.  Ans.  f r. 

9.  Show  that  the  radius  of  the  base  of  the  cylinder  of  greatest 
lateral  surface  which  can  be  inscribed  in  a sphere  of  radius  r is  \r  V2. 

10.  Find  the  shortest  distance  from  the  point  (2,  1)  to  the  parabola 

y2  = 4x.  Ans.  V2. 

11.  Determine  the  area  of  the  greatest  rectangle  which  can  be  in- 
scribed in  a given  triangle  whose  base  is  2 b and  altitude  a.  Ans.  \ab. 

12.  If  the  shape  of  a window  is  a rectangle  surmounted  by  a semi- 
circle, and  if  the  perimeter  is  given,  show  that  the  maximum  light  is 
admitted  when  the  height  and  breadth  of  the  window  are  equal. 

13.  Determine  the  altitude  of  the  minimum  cone  which  can  be 

circumscribed  about  a given  sphere  of  radius  r.  Atis.  4 r. 

el.  calc.  — 7 


98 


ADDITIONAL  EXAMPLES 


EXERCISE  14.  PAGE  64 

Prove  the  following  expansions  : 

5.  esinx  = I + x + zx2  — % x4  — T\x5  + — . 

jj-3  ^*5 

6.  arc  tan  .r  = .r  — 1 !-•••. 

3 5 7 

x 2 5 ar4 

7.  sec  X=  I + , (-  -r. — I . 

ip  [4 

X 2 X4  X6 

8.  log  sec  x = 1 1 b •••• 

& 2 12  45 

2 

9.  ez sec x = i + x + x2+ b — . 

3 

10.  sin  ( i -b  2 x)  = sin  i +2  cos  i • x + 2 sin  I • x2  + •••. 

11.  log  (I  -IB?1)  = log  2 + l X + | X2  - •••• 

12.  Vi  + ^x  + 12  x2  = I + 2 x + 4X2  + ■■■. 


EXERCISE  16.  PAGE  75 

Prove  the  following  integrations  : 

6.  ^ cos2  xdx  = \ x + \s\n2  x + c.  (Put  cos2.r  = |(i -b  cos  2 

7‘  J|f$=log(*  + *8)2  + c- 

8.  (*'*  ^X-  — x — $ x2  + $xs  — log  (x  — i)  + c. 

J x - b i 

Hint.  Divide  numerator  by  denominator  before  integrating. 

9‘  f(*  ax2  + 4bx3)*(2  ax  + 4 bx2)dx  — 1(3  ax 2 + 4 Ar3)7  + c 


10 

11 

12 


• f 
■ f 
■f 


bar*  dx  = 


-a21  + c. 


2 log  a 

■ j* cos  (mx  + n ) dx=  ~ sin  (tnx  + n)+  c. 


xdx 
V 4 — x‘ 
dx 


s~- 


m 

— \ arc  sin  x2  + c. 

2 = arc  tan  (x~  r)  + c- 


13. 

2 x + x2 

Hint.  Write  denominator  in  the  form  1 + (x—  i)2. 


ADDITIONAL  EXAMPLES 


99 


14‘  I 


dx 


V3  -4x- 4 x2 


j arc  sin  - 


. 2X  + 1 


+ c. 


15 


• Sx2  + x _ 2 dX==  l0g^  ~ l^X  + 2^8  + C' 


16 


'I 


x2  dx 


(x  + 2)'2(;tr  + i)  X + 2 


+ log(jr+  1)  + c. 


EXERCISE  17.  PAGE  79 


8.  Find  the  area  bounded  by  y=  9 — x'2  and  the  axis  XX'.  Ans.  36. 


9.  Find  the  area  bounded  by  y — 
to  ;r  = 8. 


x 

1 + x2 


and  XX'  from  the  origin 
Ans.  \oge  v'65  = 2.087. 


10.  Find  the  area  bounded  by y2  = 9 x and  y = 3 x.  Ans. 

Hint.  The  area  required  is  the  difference  of  the  areas  bounded  by 
the  individual  curves,  the  axis  XX',  and  the  ordinates  at  their  points  of 
intersection. 


11.  Find  the  areas  bounded  by  the  following  curves. 


(a)  y = x2  — 4 and  XX'. 

Ans. 

iof. 

(6)  y = 8 + 2x2-x*  and  XX'. 

Ans. 

29xf- 

(r)  y2  = g x and  x — 4. 

Ans. 

32- 

(d)  xy  =4  and  x + y = 5. 

Ans. 

6.1 14. 

12.  Find  the  volumes  of  the  solids  of  revolution  generated  by  revolv- 
ing around  XX'  the  following  areas  : 

( a ) Of  Example  8. 

Ans. 

2595  *-• 

(b)  Of  Example  11  (a). 

Ans. 

34fs  7T- 

(c)  Of  Example  11  (c). 

Ans. 

72  IT. 

( d ) Of  Example  10. 

Ans. 

fir. 

Hint.  This  is  given  as  the  difference  of  the  volumes  of  the  two  solids 
generated  by  the  two  areas  mentioned  in  the  previous  Hint. 


ELEMENTS  OF  DESCRIPTIVE 
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AN  ELEMENTARY  TEXT- 
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By  GEORGE  A.  MERRILL,  B.S.,  Principal  cf  the 
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Elementary  Geometry — Solid.  By  James  McMahon, 

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PLANE  SURVEYING 

$3-00 

By  WILLIAM  G.  RAYMOND,  C.  E.,  member  Ameri- 
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Road  Engineering,  and  Topographical  Drawing  in 
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IN  this  manual  for  the  study  and  practice  of  surveying  the 
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general  method  is  given  first  and  afterward  the  details. 
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% The  first  part  describes  the  principal  instruments  and  deals 
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SCIENTIFIC  MEMOIRS 

Edited  by  JOSEPH  S.  AMES,  Ph.D.,  Johns  Hopkins 
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groups  from  the  lowest  to  the  highest  ; and  a substantial 
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